Physics, Chemistry, & Maths Lomonosov RussiaPhysics.Word problems.8 cells.(Option for writing in MIEM).In all the proposed objectives, it is desirable to bring the answer except some, possibly convincing arguments in favor of this answer.1. a).Subway train arrives at the station "Kirov".What is the half open door before?b).At the moment there is a train stop push.In what direction?2. Evaluate the tire pressure of a car and a bicycle.Allowed to make a mistake 10 times.3. Small Chriqui thrown from the plane.He falls to the horizontal plate and an elastic rebound.What is the acceleration of the ball for the first time after the bounce (for example, when he rose from the plate by 1 cm)?4. Which of the ball more thermal capacity: that is on the table or that hangs by a thread?(Balls exactly the same).1979.Mathematics1. How many ways are the number 1979 can be written as the difference of two squares of natural numbers?2. The competition involved 10 skaters.The competition is judged three judges in the following way: each judge in his own distributes between skaters places (from first to tenth), after which the winner is the skater with the least amount of seats.What is the greatest value may take this amount from the winner (winner only)?3. At the deck of 16 cards, numbered from top to bottom.Allowed to take some of the top deck, and then removed and the remaining portion of the deck, without turning, "punch" into each other.Could it be that after several such operations the cards are numbered from the bottom up?If yes, for what is the least number of operations it can happen?Theoretical physics1. Why is extinguished match if it blow?2. Is it possible to cram 40 people into a phone booth?(Weight pers. 70 lbs.).3. In the two goals srelyayut same bullets in different places (see. Fig.).Bullets merges into the washer.Which of them will spread to the wall (without friction)?m rectangle?Mathematics oral 7-8 cells.(Game)1. By placing a circle 15 minuses.Each of the two playing with his progress can cross out one or two neighboring minus.The player who last zacherknt minus.2. 15 stones.One divides into two piles, one (and then alternate with passages) one pile takes himself another divides again into two .. The player who can not make progress.3. 15 nuts.Each of the two playing with his progress can not take yourself more than half of all nuts.The player who can not make a move.Theoretical Physics1. To determine the density of Hg (mercury) with an accuracy of 1%.For a while all the necessary data many times a day were broadcast on radio and TV.2. Will there be enough energy produced by the power plant to evaporate the water passing through the turbines it?3. Peter 2 times wider Wani, but a 2-fold lower.One of them weighs more?4. When testing a missile mounted in the tail of the aircraft to protect against attacks from behind, was discovered a surprising fact: when starting a shell he turns around and catches up with the plane.How to explain it.Experimental Physics1. Why does not take off?

2. The tape from the reel.ThanHis, thevwinderanymore.

3. Why is rotated in the fall and how to make to rotate in the opposite direction?

Chemistry1. To determine the surface area of the activated carbon.C= 4 * 103J / (kg *OC) L= 2.3 * 106J / kgC= 1 kcal / (kg *OC) L= 500 kcal / kg2. What fundamental differences should be expected in the composition of lunar minerals compared to the earth.3. How to light the burner without matches at the hardware store?4th Tournament them.Lomonosov1981.MathematicsGrade 61. From the results ofAdivided by 2;Ais divided by 2;Ais divided by 12;Ais divided into 24three true and one false.What is it?

2. To write the numbers 0;1;0;0. In one step is allowed to add one to any two of them.Can I get that all the numbers become equal?3. Several boxes together weigh 10 tons.Each of not more than one ton.How trhtonok certainly enough to take away this burden?Grade 71. Motorist rode the 4-wheel drive with a spare tire 5000 miles, changing the tires so that they wore out the same way.How many kilometers passed each tire?2. Uvadratnaya area 100 m * 100 m square is lined with slabs of 1 m * 1 m four colors.No two plates of one color are not common side or top.How many plates can be one of the colors?3. Several boxes together weigh 10 tons.Each of not more than one ton.How trhtonok certainly enough to take away this burden?8-9 classes1. Can there be a trapezoid 4-gon with sidesL= 10, 11, 12, 13,2. Dana repeating decimals with periodN.Can this period when multiplying the fraction by 2 or its division by 2a) increase;b) decrease?3. 50 at the round table.They give 101 ball, among them 2k+1 black.First 3 gets the ball, the other by 2. Then the motion of the balls according to the rules: if a person has a stroke after 3 balls of the same color - the game is over, if different - reserve the 2 identical, and gives one a different color neighbor on the right.What is the maximum number of steps can be in the game ifa)k= 1;b) 1 0), but less than half.The player who can not make the next move (this happens when the remaining two nuts).Who wins the game with the right - beginner or the second?5. The hour hand stands at 12. Two igrayuschihpo queue move her.You can move the arrow to 2 or 3 hours in advance.The winner is the one who put it on 11. Who wins the game with the right - the first or the second?6. Written number 1000000. The first player writes natralnyh products of two numbers greater than 1, equal to 1,000,000 Second selects any factor, and underneath it says equal to the product of two factors.And so on. D. In turn.The player who was unable to make the next move.Who wins the game with the right - the first or the second?(Option: the winner is the one who could not make the next move).Home1. 25 matches.Take turns, each Statement 1, 2 or 3 matches (light version 1 or 2).The winner is the one who at the end of the game will be an even number of matches.Who wins: beginner or the second?2.A- natural.The two take turns playing smash all of the available range (the first time during the whole set - the number ofA) to natural or long-terms 1 natural factors.In one move, one number is divided by 2 terms or 2 factor.The player who can not make the next move.Who wins the game if properly?a)A= 10b)A= 11Theoretical Physics6-7 class1. Boy Loew boating and water photographed beach.When he published the photos, I could not understand where the image bank, and where his reflection in the water.Specify the number of possible ways to do this.

2. Flatland - a flat country.It has only two dimensions.How can I attach the wheel to the car in Flatland?(Make-axis, as in the three-dimensional world, it is impossible - she has nowhere to hang around there).3. In carrying out accurate weighing scales placed in a hood, from under which evacuated.Why do it?4. Mikhail Lomonosov issedoval movement of air in mines complex profile.In what direction will show through the winter and summer in the mine, the illustrations?

8-9 class1. Boy Serge watched TV and decided to photograph it interesting image.When he published the photos, I saw them on the light and dark bands.How Come?What can we say about the device gate?[task involves, but has been withdrawn.]2. Towards each other fly two identical dumbbells.They elastically collide.How will fly dumbbells after the collision is over?Consider the case ofv1=v2andv1= / =v2.

3. The load falls on a vertical spring.At what height it will have a top speed?

4. In the foam float hidden two pieces of lead.What sluagh float will:1) sink:2) protrude from the water;3) to tip over the surface of the water?foam 15, 2, 2, 1,15 -> 10, 2, 2, 1,10 -> 3, 3, 2, 21 After the exchange of rubles.25 kopecks.Copper Copper determine what was silver.(Note: in 1983 in the Soviet Union were in the course of the coin 1 rub. 50 kopecks., 20 kopecks., 15 kopecks., and 10 kopecks. white, conventionally called the silver and coins 5 kopecks., 3 cop., 2 kopeks. 1 cop. yellow, conventionally called copper).3. (8 cells.) In 1983 the plane point and a circle of radius 1. Prove that there is a point on the circumference, the sum of distances between all of these points is not less than 1,983 in 1983.4. (9 cl.) Two circles intersect line.Prove that the angleABC= angleDEM.

Theoretical Physics5-7 cl.1. When will emerge from the bath faster one and the same volume of water - when it is a man or when it is not?2. On the popular lecture said that the height of the Eiffel Tower 302 meters 15 centimeters. "Well, no," thought one of those present.Why he thought so?3. Archimedes had to lift a heavy load, and had two arms:He thought - "for us, weightlifters, it's just right."Which lever he chose?8-9 cl.1. Task 1 for 5-7 cells.2. Task 2 for 5-7 cells.3. How to determine the weight of the boat, having a coil of rope?It takes place during the boat ride.4. When you turn on the speakers in the sequential translational network open one disables all in parallel - one short izamykanie disables all.How to include?Experimental Physics (5-9 cells).1. (5-7 kl.) How to measure the diameter of the thread usual line?2. How to weigh a book that weighs about 2 times larger than the limit dynamometer?3. On the inclined plane rolled 3 bubble from the glue - empty, half-full with water and.Why half-stops before?(8-9 kl.) Which part of the energy is spent on friction bubble?4. Those bubbles are.With a weak push the base falls empty, with an average - complete.Why such a sequence?5. (8-9 kl.) If narisovanna star * look through the slit, cut with a razor in the paper, the visible beam 1-2.If you rotate the gap, the rays begin to spin.How Come?Chemistry1. The ideal syringe needle is lowered into the water and pull the piston.How does the rate of filling the syringe on how much pushed the plunger and how quickly he pushed?Explain your answer.2. Can be a simple substance molecule polar?If not, why not, if so, give examples.3. As a fairly well-equipped laboratory light the alcohol lamp, if you forgot your home matches?Come up with more examples.4. In the last century there was much debate about the structure of the benzene molecule, which has a composition of C6H6.Try to draw all of the structural formula, which can describe the composition of such a molecule.Note that the C atoms can be connected double and triple bonds.5. Why concentrated hydrochloric acid fumes in air, and concentrated ammonia solution - no?6. It is known that solutions of iodine in water and alcohol have a brown color, and iodine vapor and carbon tetrachloride solution - purple.What could you explain this fact?7. When bubbling xenon, methane under a pressure of 1 atm.through water at 2OC precipitate colorless crystals of Xe * 5.75H2O and CH4* 5.75H2O, which easily decompose with the release of the gas to under heating.What do you think, what is the structure of these compounds?8. Can one get two acid salts of the two acids - salt, two bases - salt, the salts of the two - base.If the process is possible - Write the reaction - if not, try to justify their opinions.9. How does the reaction rate on the concentration and temperature / ceteris paribus /?How could you explain this relationship?10. Some waste production is oil containing water.It turns out that this oil burns even better dry.How can you explain this and how to do it?11. What, in your opinion, different substances corresponds to the following formulas: FHH | | | Cl-CH Cl-CC-Br | | | Br FFPlease note that due to carbon directed corners of a tetrahedron.Draw the molecule thereof.12. What are the inorganic substances in living organisms and why are they needed?13. 1.3 liters made of 130 g of a clear crystalline substance, which is in the process of dissolution zhidgosti level has not changed, but its temperature is lowered.After complete dissolution of the substance liquid is evaporated, with remaining 0.13 g of solid residue.Determine what kind of fluid and a substance dissolved in it.Competition in MathematicsTasks1. On the sides of the hexagon was recorded six numbers, and each vertex - a number equal to the sum of the two numbers on the allied side.Then, all the numbers on the sides and top of a single number erased.Is it possible to restore the number on the top?2. Vertices A, B, C of a triangle are connected to points A1, B1, C1, which lie on opposite sides (not top).Can the midpoints of AA1, BB1, CC1lie on the same line?3. Three chess players A, B and C played a match tournament (everyone with everyone played the same number of games).Could it be that by the number of points won first place A, C - last, but by the number of victories, on the contrary, A finished last, C - the first (for the victory is awarded one point for a draw - a half-point)?Problem solving in mathematics competition1. Yes I Am.Assume that a number of1, a2, a3, a4, a5, a6are recorded on the ribs 6 in order of the polygon counterclockwise.Then, six-sided polygon vertices will be recorded (counterclockwise) of the following:A1+ a2, a2+ a3, a3+ a4, a4+ a5, a5+ a6, a6+ a1.We can assume without loss of generality that the only erased in the top number - is a1+ a2.Then it can be recovered by the formula:(A2+ a3) - (a3+ a4) + (a4+ a5) - (a5+ a6) + (a6+ a1).After removing the parentheses it turns out that all the numbers except a1and a2, enter the right pane, double, and with different characters.2. The middle of said segments lie on the respective average lines of the triangle (as an average parallel to the base line).They also lie inside the triangle (point A1, B1, C1can not celebrate the vertices).Since the line can not cross the three sides of the triangle, the midpoints of the segments and can not lie on the same line.3. Yes I Am.Place in the match-tournament determined by the difference between the number of wins and the number of lesions.Here is a table of the tournament in 8 circles (as in match-tournament contenders in the 50s and early 60s) that meets the specifications (in the row corresponding to the player and the specified number of victories over the other players):ABCglassesvictory

A128.53

B0484

C237.55

Competition for mathematical gamesTerms Games

1. Honeycomb (1x1 squares) are arranged in a 9x9 square.In the central box instead of honey - tar.In one move is allowed to make vertical or horizontal incision (squares on the sides) and eat one of the cut pieces, in which there is no tar.Play two of the moves made in turn, loses the one who has to eat tar.Who's winning?(You can consider the case where the tar is not in the center.)2. On a strip of size 1x8 in the 4th, 6th and 8th cells are chips.Each move any of them can be pushed to the left by any number of cells, so as not to jump over another chip and not get to the occupied space.Played by two, take turns.The player who can not make a move.Who's winning?3. The arrow on the clock shows twelve hours.In one move is allowed to move in a clockwise direction to two or three hours.The winner is the one who put the arrow on the eleven.Who will win?4. The cake is in the shape of a regular pentagon.In one move is allowed to make an incision along the side or diagonally.The winner is the one whose turn is after the cut a piece containing cherry, located in the center of the cake.Who will win?Consider a plywood table on which is painted a regular pentagon in the center of which stands a vase, and the loser is the one whose turn after vase falls.5. In a series of written seven minuses.In one move, you can fix on the plus one or minus two adjacent.The player who can not make a move.Who will win?Consider the eight minuses in a row;daisy with eight or nine petals.The story of what math gamesIn this section, we did not give solutions of games offered at the competition.Otherwise, you will not be interesting to play them.But it is useful to discuss what amathematical game.With conventional games like chess, checkers, or of them have in common is that there are two players chasing opposing goals.A difference is that chess would be a fatal blow, if it turned out that the white, for example, with the right game always win.A mathematical games just interesting to invent the correct ways to play.Mathematics, however, are able to prove that the right way to play there in the ring games.But to find the way to chess and Go is not yet possible, even using super-computers (for checkers situation is somewhat different).But those games that we'll play a lot easier, and so they manage to find out who wins the game at the right.And how do understand what the right play?After a sequence of moves in the game alternates.Suppose for example, I wrote down the right, winning sequence of moves for me.After all, the enemy is not obliged to follow it!However, if it is possible to formulate a rule, following which one player always gets a prize at all possible moves of the second, it is clear that the outcome of the game is defined (it should be, of course, know this rule).Mathematics in such cases we say that the first player hasa winning strategy.It is the search of a winning strategy and one of the players is a mathematical game.Of course, if the rules of the game you like, you can it some time to play with his friends, but after a while you will notice that those wishing to participate in the game with one of the parties is getting smaller.And then, when the strategy will be known to all, the game play will not be none.Let's look at some examples of mathematical games.Task 1.There are two piles of stones, 8 pieces each.Two persons play the following game: in one move is allowed to take a handful of any number of stones.Who can not make a move - lost.Find a winning strategy for one of the players.Before embarking on the search strategy, we advise you to play this game with someone else, or at least one of the two players at once.If you have formed a definite opinion about who wins in this game, try to formulate a rule that should guide the player, ie,strategy.After that, read on and compare their solution with ours.Note that each turn number of stones in one of the piles decreases.So the game can not last indefinitely, no matter how the players play.Solution 1.The winner is the second player.To do this, he must adhere to the following strategy (called "symmetric strategy"): after each of the first player moves need to equalize the number of stones in a pile.Why such a strategy is called symmetric?Not only in this game, but also in many other positions, some are symmetrical.So, the symmetric strategy is that each turn a player achieves a symmetrical position.In this case, the position symmetrical, if the number of piles of stones in the same way.So check that the symmetric strategy is the solution of the game and provides a second player win.The first question is whether the player will always follow it?To answer this question, think about the position in which the second player can not equalize the number of stones in a pile?Only if the stones in a pile ALREADY equally.But this situation to move the second player can not occur!Indeed, initially have a symmetrical position.The first player to take a certain number of stones, and the position becomes asymmetric.Now the second question: whether the second player wins, sticking symmetric strategy?As already mentioned, the number of stones in the pile is constantly decreasing.The course can not be done when there are no stones.But this position is symmetrical, and there she was after the move of the second player.Objective 2.There are a handful of stones 4 to 9, 10, 12 and 18 stones respectively.At its course the player can take one of a handful of 1 or 2 stones.Who can not make a move - lost.Determine the outcome of the game (ie, to find a winning strategy for one of the players).This task is more difficult to last.However, try to solve it, and then read the solution given here.Solution 2.The situation in which it is impossible to make progress in this game is also unique stones --- all players.It turns out that the first player can make sure that he took the last stone.A winning strategy for the first player: do such a course, the number of stones in each pile divisible by 3.Is it always the first player can follow this rule?Note that on the first turn possible: take one stone from the heap, in which 10 stones.The second player has to take its course from one of a handful of 1 or 2 stones.If its course in every pile of stones was the number divisible by 3, and if the first player follows the preceding rule, this is true, then before the next move of the first player in exactly one pile of stones is not the number is divisible by 3. It is clear that the first player can follow the preceding rule, taking from this heap remainder of dividing the number of stones on the 3.So, after the move of the first player the number of stones in a pile to be divided by 3. But the game ends when the stones in piles - 0 0 0 0 - iethe first player wins.Competition in PhysicsAssignmentsLower grades (6-7)1. Why icicles are usually located only one side of the roof?2. Air and water are transparent.Why is opaque fog?3. Sergei Archimedes believes that he is lucky with good weather.More than once he noticed that when the clouds surrounded the entire horizon, namely the sun shines on it.Does Serge rights?Upper grades (8-9)1. Sergei Archimedes believes that he is lucky with good weather.More than once he noticed that when the clouds surrounded the entire horizon, namely the sun shines on it.Does Serge rights?2. Wider or narrower than the moon moonlit path on the water (meaning angular dimensions)?3. Rats and other rodents some teeth stay sharp all the time, despite the constant stitching.This is because their teeth consist of materials with different hardness.Explain how such self-sharpen your teeth and how they are distributed substances with different hardness.Competition in AstronomyQuestions1. When people built the oldest observatory?Why is it needed?2. Eratosthenes in the 2nd century BC, knowing nothing about other countries, to measure the size of the Earth.How is it managed?3. How to change the speed of the satellite, if it starts to slow down?4. They say: "light and not heat."About what I may say so, and why?And whether there is other way around?5. Why is the planet in the sky after the Sun go?6. To the sky wants to make all well verree great impression, and in fact quite small and nevzrachnenky?7. What constellations in the sky most ancient?Do you know why they are so named?8. Why do not we see how stars are born?Can we see how they die?9. When humanity begins to travel to other planets that we will be interested in it?10. What are the similarities between the universe and the cotton candy?Answers to questions of competition on astronomy1. The answer to this question depends on what is meant by the observatory.A.If this religious building, at least to some extent related to astronomy, the oldest known structure - Stonehenge - the remnants of a giant megalithic stone structure, built at the turn of the Stone and Bronze Ages (1900-1600 BC) on the territory of modern England.At Stonehenge carried out not only ritualistic ceremonies, but also, due to the special orientation of the individual parts of the structure, was conducted through calendar days, marks the beginning of the seasons.B.If you have in mind for the construction of regular visual observations of stars, it is an observatory of ancient Babylon (1st half of the 2nd millennium BC).There were widespread predictions of important events occurring on the basis of the heavenly signs, and thought: the more precise observations, the more certain prediction.It appeared in Babylon associated with astronomical events mathematical time-scale (8th century BC).B.If we talk about observatories equipped with tools, then we should talk about ancient Greece, about the age of antiquity.

2. Eratosthenes lived in Alexandria in northern Egypt.And in the south of Egypt was the city of Siena, which was a wonderful well.He was vertical and deep, and at noon, the longest day the sun illuminates its bottom, iewas at its zenith.Eratosthenes measured figured in one of those days the sun zenith angle (angle to the vertical, it was 7.2O), imagine the Earth ball, put Siena and Alexandria on one meridian and ascribe 7.2Omeridian arc distance between cities.Knowing this distance and the fact that a full circle is 360O, it is not difficult to find the proportion of the circumference of the Earth's meridian.Eratosthenes was 46,000 km.Excellent accuracy!3. If the satellite will begin to slow down, it will spiral closer to the ground.Hence, it has a circular orbit at an initial vertical velocity component to appear.If braking is slow enough, then for each new round circular orbit can be considered, but a smaller radius H. Sincecircular velocity V = (GM / (R + H))1/2(R - radius of the Earth, H - height above the earth, M - mass of the Earth, G - gravitational constant), we see that Vcrincreases with decreasing H.iewith slow growing and normal braking, and the tangential component of the velocity, and hence the full velocity vector.The growth rate is due to the completion of the kinetic energy of the satellite mv2/ 2 due to the potential energy mgH.4. "Light, and does not heat" - can be said about any source that emits in the optical range at low power.On the contrary, you can say about the powerful sources of electromagnetic radiation with a peak in the infrared region of the spectrum.Example - iron.5. The solar system is highly flattened.Planets very slightly spaced from the central plane in which they revolve around the sun, all in one direction.This plane is called the ecliptic.Therefore, the observer on Earth, and it would seem that the planets and the Sun on the background of distant stars move in the sky on the same line.6. Solution of this problem is not given, becausethe collector could not learn what it had in mind the authors.7. It is difficult to say whether the names endowed beautiful configuration brightest stars Adam and biblical patriarchs.Ancient texts mentioning the constellations date from the second half of the second millennium BCPerhaps the oldest known - Old Babylonian text (1700 BC).This is the prayer to the starry gods.Referred to the constellation - participants myths (monthly myths, according to the lunar cycle).Obviously, the first in this regard, attracted the attention of circumpolar constellations, especially the Big Dipper and the North Star in Ursa Minor, because the rotation of the sky takes about an axis passing near the North Star.These constellations are the first to receive sustainable titles (based on mythological), which in most cases are not associated with modern, inspired later periods (Antiquity, the Middle Ages, the era of geographical discoveries).8. Stars are born when compressed gas and dust clouds, with a sufficiently high degree of compression when warming up in the center is sufficient to "ignite" thermonuclear reactions.Death of a star occurs when the star is no longer in a stable condition under normal parameters (density, size, somehow comparable with solar).In this sense, the stars in store for several kinds of deaths, according to their initial weight.When M RCOOH+3R'OH

MR(x) 3M +R '(3Y) +6036x + 453y + 51

7.423.226.72

0.14 mol0.07 mol0.21 mol

We hope that a solution to this proportion did not take long at any of the aforementioned scholar nor a physicist, and they will find that MRacid - 46, alcohol - 32, and saponified orthoester of formic acid - HC (OMe)3.3. Again - the logic does not work honors.The calculation for ordinary metals A and B gives the answer, not provided for periodic system:the mass fraction of hydrogen24 / (a + b + 192 + 216) = 0.0453a + b = 121,8

the mass fraction of oxygen320 / (a + b + 192 + 216) = 0.5176a + b = 210,3

4. (Atomic mass A and B taken as a and b, respectively).And it does not happen.Hence, not only the hydrogen contained in water of crystallisation.5. It is assumed that one of the "metals" - not metal, but something containing either oxygen or hydrogen (in this case rectilinear invalid payment).Scrabble recalls that happens also ammonium, behaves quite "metallic", a physicist calculates it logically.Therefore, ammonium alum.If a = 18 (ammonium), then b = 210-18 = 192 - is iridium Ir.Formula alum -NH4Ir (SO4)2* 12H2O.Well, those who are afraid of words that are not in the school curriculum, it would be better not to go to the Olympics (especially since there is nothing new in the properties of iridium with respect to aluminum casting is not required).6. We form the equation for the reaction:CaO + H2O = Ca (OH)2;56 g CaO react with water to form a 74 g Ca (OH)2,CaO A g react with water to form (74/56) A g Ca (OH)2Thus, in the solution will contain 1.32 g A Ca ( OH)2Since the weight of the solution (on the basis of the reaction equation) is equal to the sum of the weight of the starting materials and (A + B) r, starting material is completely soluble (no precipitate) have= 132 * A / (A + B) (%).7. We form the equation for the reaction:CuO + H2= Cu + H2O;1 mol of CuO - 1 mol Cu,80 g / mol - 64 g / mol,80 g of CuO give 64 g of Cu.

Hence, the mass of the sample taken after passing over it hydrogen decreases in 80/64 = 1.25 times.8. Methods color celluloid based on his restoration various nitrogen-containing compounds.On this and many other interesting things in organic synthesis is well written in the book Shpaustsusa "Journey into the world of organic chemistry."

Tasks1. (7-11) on the plane is a square, and invisible ink applied point.Man in special glasses sees the point.If you draw a straight line, he said, is whether a point on the line, and if not lies, he says, on which side of this line is invisible point.What is the minimum number of lines necessary to carry out, to see whether the invisible point lies inside a square (not outside and not on the border)?2. (7-9) Is there a 100 natural numbers such that their sum is equal to their least common multiple?3. (7-9) The rectangle ABCD (AB = a, BC = b) It so happened that turned the pentagon area S (C laid down in A).Prove that S 2NO2NO + O2-> 2NO2In the absorption solution CO32-:1) H2O + NO + NO2+ 2CO32--> 2HCO3-+ 2NO2-2) 4NO2+ O2+ 2H2O + 4CO32--> 4HCO3-+ 4NO3-Case 2) is impossible.Hence, the gas mixture before absorption - NO, NO2and, possibly, N2.After absorption - N2+ NO.Suppose, in terms of 1 mole of a mixture (0.5 mole of N2O and 0.5 mol2) to get the x mol NO2.We have:Number N: 2x + y + 2z = 1Number O: 3x + y = 1The volume ratio of y + z = 5/8Hence, y =11/20, x =3/20, z =3/40.Reacted1/2-3/40=7/40mol N2of 0.5.Hence, reacted 70% N2.7. You can use any reasonable method of analytical determination of H2, Zn2+and Fe2+.For H2- determination of redox-potential of water for Zn2+sulfide method for Fe2+- [Fe (salt)]3-.The most promising method is based on detecting the presence of lubricants, for example, using adsorbents or analyzing a surface layer of water on the presence of organic impurities.

Competition in Mathematics

TasksGrades 7-91. From the hot tub faucet filled in 23 minutes, from a cold - 17 minutes.Mary opened the first hot tap.How many minutes it should open the cold, so that when filling the bath with hot water nalilos 1.5 times more than the cold?2. The length of the height of the rectangular trapezoid ABCD AB equal to the sum of the lengths of the bases AD and BC.In what respect bisector of angle ABC divides side CD.3. At each kilometer of highway between the villages of Elkin and Palkino should post a sign on one side of which it is written, how many kilometers to Elkin, on the other - to Palkino.Boris noticed that each column is equal to the sum of all digits of 13. What is the distance from Elkin to Palkino?10-111. Any more numbers among all the numbers from 100 to 999: those in which the average number of both more extreme, or those in which the average figure is less than both the extreme?2. Circle held several (finitely many) different chords so that each of them passes through the middle of any other of the studies chords.Prove that all these chords is the diameter of the circle.3. Is there a convex polyhedron has 12 ribs, which are, respectively, 12 and parallel to the diagonals of the faces of a cube?Additional tasks (for all classes)1. Find the sum of angles of MAN, MBN, MCN, MDN and MEN, drawn on graph paper as shown.2. 3. * -ABCDE4. | | | | | |5. * - * - * - * - * - *6. | | | | | |7. * - * - * - * - * - *8. | | | | | |9. * - * - * - * - * - *10. | | | | | |11. * - * - * - * - * - *12. | | | | | |13. MN - * - * - * - *14.

15. Dan paper circle.Can I use the scissors to cut it into several parts that make up the square of the same area?(Cut permitted under direct and arcs).16. In the bazaar selling fish, big and small.Today, three large and one small stand together as much as three large and one small yesterday.Is it possible to find out from these data that the more expensive: one large and two small today, or five little yesterday.17. Peter cut a rectangular piece of paper in a straight line.Then he cut in a straight line from one of the resulting pieces.Then he did the same with one of the three pieces of the resulting etc.Prove that, after a sufficient number of cuts can be selected among the 100 pieces of the resulting polygons with the same number of vertices (e.g., triangles 100 or quadrangles 100, etc.).Problem solving in mathematics competitionGrades 7-91.To the hot water in the bath turned out 1.5 times more than the cold, cold tap must fill 2/5 baths and cold - 3/5 (to find out, it was possible to identify the volume of the bathroom for 1, the amount of cold water for x; if x + 1,5x = 1, where x = 2/5).But if the hot tap must be open all (2/3) * 23 = 69/5 minutes, and the cold - for (2/5) * 17 = 34/5 minutes.Hence, the need to open the cold tap over (69/5) - (34/5) = 35/5 = 7 minutes.2.We prove that the bisector of angle ABC chord divides DC half.Let O - midpoint of the side DC.Draw through the points A and O line, let M - the point of intersection of this line with the line BC.Then triangles AOD and MOC will be equal to the second feature (DO = OC by hypothesis,/AOD =/COM as vertical;/= ADO/OCM, as the lines AD and BC are parallel).This means that AO = OM, t. E. The triangle ABM - isosceles.But bisector at the vertex of an isosceles triangle is the median and the height from which the bisector of angle ABC passes through the middle of AM, that is, the point O, and then divides DC into equal parts (ie. A. O - mid-DC).

3.Let the distance from Elkin to Palkino n kilometers (by hypothesis, n-integer).We number the columns from Elkin to Palkino order.Consider the 9th column, t. E. A pillar, spaced from Elkin 9 kilometers (clearly, n>10).Then, one side is written 9, and on the other: n-9.The next column is written on one side 10 and the other: n-10.If n does not end at 9, then the n-9 would not have ended with at 0, and hence the amount of digits of n-9 and n-10 would be equal, but then on the 9th column sum of the digits would be 8 more than on the 10th, which is impossible.Hence, n ends at 9.If n> 49, then the sum of the digits in the 49th column is greater 13. This means that for n is the only such features: 19, 29, 39, 49.If n = 19, then on the 9th column sum of the digits is equal to 9 + 1 + 0 = 10 - a contradiction.If n = 29, then on the 9th column sum of the digits is equal to 9 + 2 + 0 = 11 - a contradiction.If n = 39, then on the 9th column sum of the digits is equal to 9 + 3 + 0 = 12 - a contradiction.There is only one possibility: n = 49.It is easy to check that in this case, the sum of all columns of numbers will be equal to 13 (it is sufficient to check only for 9, 19, 29, 39, and 49 poles, think why).10-111.Note that if a number greater than the average number of both extremes, then the number of 999-a figure less than the average of both extremes.Therefore, some of the numbers from 100 to 999-100 = 899 the same amount of numbers with the highest average figure and the lowest average number (which can be divided into pairs).But there are still a number from 900 to 999. It is clear that none of them are numbers with the greatest average number, but there is the lowest, such as 901. Therefore, greater numbers of those who have less than the average number of both extremes.2.Note that the smaller the distance from the center O to the circumference of the chord, the greater the chord length.Since the final number of chords, among them is the smallest in length, for example, AB.By hypothesis, it passes through the midpoint of chord K of some other, say, CD.If the point of intersection of AB and CD is not well and the middle of AB, the distance from point O to CD will obviously be greater than the distance from O to AB (t. To. OK will be greater than the length of the perpendicular from the point O to AB) therefore, the chord has a smaller CD than AB, length - contradiction.So, CD goes through the middle of AB, where the perpendicular from the point O on these chords are the same.This is possible only if the AB and CD are the same, or if the AB and CD intersect at the center.The first is impossible, then AB and CD - diameters.Likewise, it is proved that all the other chords pass through O.

3.The answer is: there is.Example.Construct a cube whose edges are twice more than the initial, middle note and its faces.It is clear that one can construct an octahedron with vertices at the marked points, with all its edges are equal halves of the diagonals of the cube faces constructed.Thus, this is the desired octahedron.Additional tasks (for all classes)1.Note that/= MAN/RES,/= MBN/QER,/MCN =/PEQ,/MDN =/NEP.Hence we see immediately that/MAN +/+ MBN/MCN +/+ MDN/MEN =/MES = 45o

2.Let us assume that this can be done.Then the edge of each piece is made up of lines and arcs of circles.For each arc define its "angular measure", equal to the ratio of length to the radius of the arc, with taken with a positive sign if the radius vector drawn from the center of the circle, sticks "inside" our paper circles and negative if it sticks in outside.Now we assign each piece of the amount of angular measures its arcs.It is easy to see that if we "sostykum" together two pieces A and B, the sum of the angles measures piece A + B is equal to (the sum of the angular piece measures A) + (sum of the angles measures piece B).This is because when two arcs are glued, the amount of angular steps equal to 0.Hence we see that the sum of the angles measures of the arcs on the circle should ovpadat with the sum of the angular measures of the arcs on the boundary of the square.But this is not the case for the circle is equal to the number 2, and for the square - 0.3.Let 'fish prices ": today the big fish worth bC, and small mC.Yesterday most cost bv, and the little - mv.Then from the conditions of the problem we have two equations3bCm +C= 5bv, 2bCm +C3b =vm +v.Hence, we obtain:5mv= (2bCm +C-3bv) 10b 5 =C+ 5mC-3 (3bCm +C) = bC2m +C.That is five little yesterday cost as much as one large and two small today.4.Note that when Petya dissections obtained only convex polygons.In addition, we note that the cutting convex n-gon in a straight line, we get da new polygon with n1and n2and n vertices, respectively1+ n23 * 99 + 4 * 99 + 5 ((3 * 99 + 1) -99-99)(Ie. A. Of the numbers n1, ... is at most equal to 99 3 to 99 4 equal, and the remaining number is not less than 5).Then12 + 4 99 *>99 * 7 + 99 * 5 + 5therefore 4>5 - a contradiction.So, after 3 * 99 + 1 Petya's sections there are 100 polygons with the same number of vertices.

Competition in PhysicsTasks1. From the tower height H jumping stunt tied cord.The length and rigidity of the cord are chosen so that stunt stops near the ground.Made some hesitation, stuntman hangs above the earth at an altitude of L. Find the maximum speed that the stunt was during the jump.2. Water is poured into the bottle to the point where the bottle begins to narrow, and the top was poured kerosene (kerosene lighter than water and not miscible with it).The bottle is then shaken several times and put in place.How to change the pressure on the bottom of the bottle?3. Ball from pind-pong table falls from the Ostankino Tower and elastically strikes the ground.Determine the acceleration of the ball immediately after take-off.Qualitatively draw a graph of acceleration versus time.4. Narrow plate of length L lies on a rough horizontal table.Plate starts to rotate at a constant low speed by applying a force (finger) on the side at one end of the plate.Determine where the point around which the plate.5. Consider the following model of a clock.The light beam is successively reflected by two mirrors arranged parallel at a distance L. Each time the beam is reflected from the mirror 2, the electronic circuit for increasing the clock 2L / c, c - velocity of light.Show that if these clocks give some speed v in a direction parallel to the plane of the mirror, the speed of the clock is reduced, and find how many times (this is - the famous "relativistic reduction of the time").(Reminder: the speed of light in any frame of reference is the same and equal to c = 300,000 km / sec) .6. As you know, on a moonlit night on the water forms the light path (where should be a reflection of the moon).Explain (qualitatively) its shape.As it depends on the width of the wave heights (or ripples) on the water?7. Before you caliper.Determine (with his help) what color - red or green corresponds to the shorter wavelengths of light.Note:one of the first who vysskazala the idea that the color is determined by the wavelength was Euler.He believed that the red color corresponds to a smaller wavelength than the green.Come up with a simple way to check this.Was Euler rights?Problem solving competition in physics1.The speed stunt will be a maximum at a time when the maximum of its kinetic energy.The kinetic energy at any given time is equal to the work committed over stuntman external forces at a given moment.External forces that should be considered in this problem are the force of gravity and the force of the elastic cord.Let the cord length is l, stiffness k, mass stunt m, the gravitational acceleration g.If we denote by x the distance from the top of the tower, which is flying stunt to this point in time, the work force of gravity at this point is mgx, and the work of an elastic spring -kx2/ 2.Work elastic force is negative, since this force is directed against the movement stunt.We obtain the expression for the kinetic energy stuntmv2/ 2 = mgx-k (xl)2/ 2 = - (k / 2) (x - ((m / k) g + l))2+ (m2g2/ 2k) + mglThe value of this expression is maximal and equal to (m2g2/ 2k) + mgl, when (x - ((m / k) g + l)) = 0, i.e.m (vMax)2/ 2 = (m2g2/ 2k) + mgl.From WherevMax= ((m / k) G2+ 2GL)1/2.It remains unknown to us here to express the magnitude of the problem through data.To do this, we rewrite the conditions of the problem in the form of equations containing these unknown quantities.What stunt stops at the ground, means that the work of the external forces is equal to 0 when the tensile cord still Hl, ie0 = - (k (Hl)2/ 2) + mgH.That stunt eventually freezes at a height L from the ground, means that the elastic force of the pinch adjustment equal to the force of gravity, i.e.k (HLl) = mg.For computational convenience we denote u = k / m.Then our terms and conditions can be rewritten as two equationsu (Hl)2= gH, u (HLl) = gwith respect to two unknown l and u.Solving this system and substituting the solution into the expression for the maximum speed, we get the answer:vMax= ((G2/ u) + 2GL)1/2= G (H + L-(2H (L (H / 2)))1/2).2.Choose a column of liquid from the surface to the bottom of the bottle.Shaking up to a column made up of clean water column and the column of kerosene over the water column.Pressure on the bottom after shaking changes so as to change the weight of the selected column, which at the same size will now consist of water and kerosene.So, find out the column will be easier or harder.First consider that a very narrow neck.Then the kerosene fraction in a mixture with water is very small, and the weight of the column of our practically become equal to the weight of the column of pure water.This column will be so severe initial composite column as the column weight of kerosene less than the weight of the column of water, which replaced it in the neck after shaking.Hence, the bottom pressure will increase.Then it remains to note that the column weight of a mixture of kerosene and water can be equal to the weight of the composite column to shake only when the width of the neck of the bottle becomes equal width.3.The ball is very light, so you need to take into account air resistance to its motion as well as it is necessary for the fall in the air a feather light.As the air resistance increases with increasing speed of the object in the air, then falling with a high tower, the ball eventually attains a speed at which the air resistance force becomes equal to the gravity of the earth ball - its weight.After that, the speed of the ball stops changing, because the sum of forces acting on it is equal to 0. With this speed and it will reach the ground.The ball is very elastic, therefore, after hitting the ground, the magnitude of its velocity will not change much, but the speed will be directed up now.The force of air resistance, remains the same in size, that is equal to the weight, will be directed down now.At this point the value of the sum of the forces applied to the ball, will be twice the strength of his attraction Earth, and thus the acceleration of the ball is twice the gravitational acceleration g.4.We assume that a moving plate portion of a fixed length s of the friction force acting ks, where k - coefficient of friction which is independent of the position on the plate portion, that is the same throughout its length and is independent of the velocity plot.Then, in order to maintain uniform motion applied finger force F must uravnoveshivapt total force of friction:F = kL.That the movement is a uniform rotation around a fixed center means that no external force with respect to this center is 0. Let l be the distance from the pin to the center of rotation.Then the value of the moment of the force F is equal to Fl = kLl.We calculate the friction torque acting, for example, on a segment from the center to the end plate, which acts on the finger.Coordinates of a point on this segment can be regarded as the distance from it to the center of rotation.We divide the interval into N small plots points with coordinates x0, x1, ..., xN(in this case it turns out that x0= 0 and xN= L) and add the value of the friction torque acting on these sites .The magnitude of the frictional force applied to the portion of xito xi + 1, k is equal to (xi + 1-xi).Distance from the center of rotation to the point of application of this force can be approximately regarded as the distance to the middle of the plot (xi + 1+ xi) / 2.Then the value of the friction force at the moment of this portion is equal to k (xi + 1-xi) (xi + 1+ xi) / 2.Adding points in all areas, we obtain((1/2) k (xN)2- (1/2) k (xN-1)2) + ... ((1/2) k (x3)2- (1/2) k (x2)2) + ((1/2) k (x2)2- (1/2) k (x1)2) + ((1/2) k (x1)2- (1/2 ) k (x0)2) == (KL2/ 2) - (k02/ 2) = kl2/ 2.Likewise we find that the magnitude of the friction torque acting on the segment on the other side of the center is equal tok (Ll)2/ 2.Given the direction of the forces, we write the condition for the vanishing of the sum of the moments of the external forces0 = Fl - (kl2/ 2) - (k (Ll)2/ 2) = kLl - (kl2/ 2) - (k (Ll)2/ 2).From this equation we find that the distance from the finger to the center of rotationl = (1- (21/2/ 2)) L5.Let us find the timebetween two successive reflections of the beam from the mirror to the observer 2 K, which sees our model moves with velocity v parallel to the plane mirrors.To do this, the length of the path traveled by the beam between the reflections, divided by the speed of its movement.The speed of the beam in the reference frame K (as in any other frame of reference) is equal to c = 300,000 km / sec.It remains to find the path length (see. Figure), which is equal to A1B + BA2= 2A1B.You can write the equationA1B2= A1C2+ BC2.where BC = L, A1,C = A1A2/ 2.Express the members of this equation in terms of the data of the problem and the required time.A1A2- is the distance at which moves our model in the time between two reflections, A1,A2= v.On the other hand, A1B + BA2- is the distance that light passed during the time between two reflections, A1B + BA2= c, and hence, A1with B =/ 2.Substituting these expressions in our equation, we obtain the equation for:(C/ 2)2+ (v/ 2)2= L2,from where= 2L / (C (1- (v / C)2)1/2).Thus, an observer noted K time between two reflections='/ (1- (v / C)2)1/2.where'= 2L / C - the time between these two events marked by an electronic circuit, for which our model is stationary.This means that the speed of the clock, driven by an electronic circuit, after imparting a velocity model v, will be less than the clock speed observer ostavshekosya stand still in/'= 1 / (1- (v / c)2)1/2times.

6.If the water surface was mirror smooth, no track, we would not have seen.Water would create a clear reflection of the moon, visible in one specific point on the surface of the mirror.If we see a track, this means that the surface waves act as small mirrors which are due to the different tilt send us moon light from points that are away from the place of its reflection mirror smooth.Variation in position of the points on which falls to us moonlight, i.e. the track width, the greater the slope of a strong possibility of small mirrors, that is, the higher the water waves.If viewed as a wave of convex reflecting surface is incident to the beam on the surface of its reflection from all points on the surface of a cone, which is wider, the more convex surface, or higher than the wave (in the case where the surface is flat, i.e. zero wave height This cone is compressed into a reflected beam).Adjudged whether reflected from the water space of the light to the observer, i.e. whether it is in place within the lunar tracks depends on whether the observer is located within the cone of rays created at that location with a maximum wave height.If we consider the maximal cones of reflected rays at different locations of the water within which enters the observer standing on the shore, it turns out that the path widens when approaching the observer.7.All that we need in this experiment from the calipers - a narrow slit that can be made between his wings.To zaintereovatsya this gap, we must remember that the intensity of the received light waves passing through the slit depends on the angle distribution after the gap.For waves of a given frequency, this dependence appears in the form of alternating-receiver on the screen (or the retina of the observer) of light and dark bands parallel to the slit.This phenomenon is called diffraction.The fact that the waves arriving at a given point from different angles slit may amplify or attenuate each other depending on the difference of the lengths of the paths that they passed.If the wavelength is much greater than the width of the gap, the waves coming from different points of cracks practically do not differ from each other and reinforce each other znachittolko regardless of the propagation angle.For an observer it looks as one broad band light.Dependence of the intensity on the angle becomes stronger, and therefore the width of the strips and the distance between them is smaller, the shorter the wavelength.Thus, if the distance between the strips of one color is less than the other, the colors have a wavelength of shorter length.And if you look at the bands of red and green colors of the light transmitted through the leaf calipers, it becomes clear that Euler was wrong.

Competition in ChemistryTasks1. Masa piece of chalk and 50 g of calcined residue was dissolved in 1 liter of water.Through the solution at room temperature was omitted chloro obtained by reacting 250 ml 20% HCl solution (= 1.1 g / cm3) with an excess of potassium permanganate?a) Which substances contained in the solution?b) Evaluate the content in it the main reaction product.c) Determine the mass fraction of calcium carbonate in the original piece of chalk, considering that chlorine has reacted completely.2. The combustion oxygen unknown substance formed 3.6ml water and 2.24 liters of nitrogen (standard conditions).Relative vapor density of the substance to hydrogen is 16. Determine the molecular formula of the substance.What properties it possesses?Where to use?3. Determine how many grams of a 10% solution of sulfur oxide (6) into pure sulfuric acid and 60% solution of sulfuric acid is necessary to prepare 480 g of a 90% solution of acid.4. For a complete combustion of the organic substances required twice less oxygen than for complete combustion of the next member of the homologous series.It can be any compound.5. Four chemical element designated by the letters A, B, C, D. Pick such reactions, which can be encrypted as follows:a) 2CA + A2= 2CA2b) D4C3+ 6B2A = 3CB4+ 2D2A3a) CB2A2+ CB4A = C2B4A2+ B2A6. Metal plate weighing 50 g after staying in a solution of hydrochloric acid decreased in weight by 1.68%, with 0.336 liters of gas stood out.From what can be made of metal plate?7. How to change the rate of reaction between the molecules of nitric oxide (2), and oxygen, if a sealed vessel filled with a mixture of these gases in a ratio of 2: 1, the pressure increase is 3 times?Problem solving competition in Chemistry1.Assuming that the chalk is pure CaCO3, we have 0.5 moles of calcium carbonate.For equationCaCO3= CaO + CO2amount of CaO and 0.5 mol amount.After the process ofCaO + H2O = Ca (OH)2obtain 0.5 g of 37 mol or lime.In 250 ml of 20% HCl solution sorderzhitsya 250 * 1.1 * 55 = 0.2 g of hydrogen chloride, which is 55 / 36.5 = 1.507 mol.For equation2KMnO4+ 16HCl = 2MnCl2+ 5Cl2+ 2KCl + 18H2Oobtain 1.507 * 5/16 = 0.471 mol or 17.19 g of chlorine.a) reaction of lime with chlorine at room temperature proceeds according to the equationCa (OH)2+ Cl2= CaOCl2+ H2OIt is evident that the lack of chlorine, and may be present in solution in addition to the main product CaOCl2and Ca (OH) Cl, CaCl2, Ca (OH) OCl, Ca (OCl)2, Ca (OH)2.b) The maximum content CaOCl2solution is(0.471 mole * 127 (g / mol)) / (100 g + 37 g + 17.9 g) = 0.0567 or 5.67%c) If reaction (1) passed quantitatively, the lime, and accordingly, the oxide and calcium carbonate was 0.471 mole.In this case, the mass CaCO3: m = 0,471 * 100 mole (g / mol) = 47.1 Then, the(CaCO3) = 47.1 / 50 = 0.942.2.The substance may comprise hydrogen, nitrogen, kislolrod.However, the relative molecular mass of the substance 16 * 2 = 32 limits the number of oxygen atoms to one or eliminates it.By the condition of nitrogen formed 2.24 liters, which is 0.1 mol.Hydrogen 0.4 mole water;0.1 mol of N2were 0.2 moles of nitrogen atoms.The ratio of nitrogen and hydrogen in the compound is 0.2: 0.4 = 1: 2.Simple formula - NH2.MR(NH2) = 16;MR(N2H4) = 32 to coincide with the defined problem statement.From the formula of the starting material H2N-NH2.This hydrazine at room temperature it is a colorless liquid that is used as one of the propellant components.3.10 g of 10% oleum contains 10 g of SO3and 100 g of 90% sulfuric acid.If the count content of SO3, then 90 g of H2SO4is contained in all (18/98) * 90 = 16.53 g of water, the rest 100-16,53 = 83.47 g oleum is SO3.Similarly, in the case of 60% sulfuric acid: 40 + 60 * (18/98) = 11,02 + 51,02 = 40 g of water and 48.98 g of SO3.In 480 g of 90% sulfuric acid contains 0.9 * 480 = 432 g of P2SO4or (80/98) * 432 = 352.65 g SO3and (480-352,65) = 127.35 g Water .Suppose we must take x g oleum, then a 60% acid - (480-x) In the oleum contains (x / 100) * 83.47 g SO3, and acid ((480-x) / 100) * 48.98 g of SO3, for a total of 352.65 g of SO3.Hence:(X / 100) + 83.47 * ((480-x) / 100) * 48.98 = 352.65.Solving the equation, we get x = 340,8 g of oleum, and a 60% acid, respectively 480-340,8 = 139.2 g4.For the oxidation of the group -CH2- (homologous difference) need three oxygen atoms.Therefore, the starting material must also be oxidized by three oxygen atoms, then the next member of the homologous series will be oxidized by six oxygen atoms.CxHy+ (x + (y / 2)) O2= xCO2+ (y / 2) H2O,Cx 1 +H2 + y+ (x + (y / 2) +1,5) O2= (x + 1) CO2+ ((y / 2) +1) H2OExamples of substances oxidized by three oxygen atoms (the first term gamologicheskih series):CH3OH - methanol,HOOCCH (OH) COOH - gidroksimalonovaya acid,CH2N2- diazomethane,HO-CH2-COOH - hydroxyacetic acid,HO-C ( COOH)3- trikarboksimetanol.5.A - oxygen, B - is hydrogen, C - carbon, D - aluminum.6.Reaction equation: M + nHCl MCl =n+ (n / 2) H2.The weight loss was 50 * 0.0168 = 0.84 g of metal x is isolated (n / 2) * 22.4 L H2, and 0.84 g - 0.336 H L2, x = (0,84 * n * 22.4) / (2 * 0,336) = 28n ofIf a monovalent metal, m. E. N = 1, it is the relative atomic weight 28 (silicon).However, silicon does not dissolve in hydrochloric acid and may be monovalent.For n = 2 the relative atomic mass is 56 tons. E. Metal - iron.You can check the options with n = 3 and 4, and make sure that the correct answer - only iron.7.Reaction equation: 2NO + O2= 2NO2.Denote the concentration of NO letter a, O2- b, then to the pressure change speed reuktsii V1= ka2b.With increasing pressure, the concentration increased by 3 times, skrost is V2= k (3a)23b = 27kab;increasing the speed V2/ V1= 27 times.Competition in MathematicsTasksGrades 7-9 (or younger).1. By the banks of the Nile came a company of six people: three Bedouin, each with his wife.At the shore is a boat with oars, which stands only two people.Bedouin can not allow his wife was without him in the company of other men.Can the whole company to cross to the other side?See comment on the condition of the jury of this problem.2. In triangle ABC, angle A is 120 , the point D lies on the bisector of the angle A, and AD = AB + AC.Prove that the triangle DBC - equilateral.3. Written in a circle 7 natural numbers.It is known that in each pair of adjacent numbers divided by one another.Prove that there exists a pair of adjacent numbers and not with the same property.10-11.1. Number of 1/42 spread in infinite decimal fraction.Then crossed 1997 th digit after the decimal point, and all the figures standing to the right of strikeout numbers have shifted to the left one.How many more: new or original?2. Is it possible to cut an equilateral triangle into five distinct isosceles triangles.3. Antique dealer bought 99 identical in form of ancient coins.He was told that exactly one of the coins - a fake - this is easier (and real weigh the same).How using pan scales without weights for weighing 7 to identify the counterfeit coin, if an art dealer does not solve any coin weigh more than two times?Problem solving in mathematics competition1. (7-9).The jury's comment on the condition of the problem 1 for grades 7-9The jury believed that the wife can not "be in the company of another Bedouin without her husband" even infinitesimal time.That is the situation when a woman swims in a boat to the shore, where there is her husband, but there is another Bedouin, and without leaving the boat sails away, was, according to the jury, contrary to the statement of the problem.However inadvertently, this condition explicitly in the final text of the problem has not got, and when checking for his failure to score was not reduced.Any other decision of the jury students could rightly challenge, since the concept of "being with" uniquely defines the above situation.Back to the condition of the problem.

We introduce the notation: B1, B2 and B3 - Bedouins, x1, x2, x3 - their wives,[}- boat.Bank of the NileNileThe other bank of the Nile

B1 + x1, x2 + B2, B3 + R3[}There is no one

B1, B2, B3 + R3[X1, x2}There is no one

B1, B2, B3 + R3{X1]R2

B1, B2, B3[X1, R3}R2

B1, B2, B3{X1]R2, R3

B1 + x1[B2, B3}R2, R3

B1 + x1{B2 + x2]B3 + R3

X1, x2[B1, B2}B3 + R3

X1, x2{R3]B1, B2, B3

R2[X1, R3}B1, B2, B3

R2{R3]B1 + x1, B2, B3

There is no one[X2, x3}B1 + x1, B2, B3

There is no one[No one}B1 + x1, x2 + B2, B3 + R3

2. (7-9).Note on the segment AD point K such that AK = AB.Then by hypothesis KD = AC.ABK is equilateral triangle, since it has two equal sides and one of the angles of 60 .Therefore, the triangles ABC and KBD are equal on both sides (AB = KB and KD = BC) and the angle between them (the angle BAC = angle BKD = 120 ).So the sun = BD and angle DBK = angle CBA.Adding to both sides of the last equality angle KBC, get the angle DBC = angle KBA.Thus, triangle DBC - isosceles 60 at the vertex, so equilateral.13. (7-9).Connect pairs of adjacent numbers so that the arrow came from the fold (so-called number that is divisible by the divisor) to the splitter (if neighboring numbers are equal, then the direction of the arrow is chosen arbitrarily).The total number of arrows is odd (7), so they can not alternate directions.Hence, any two adjacent arrows in the same direction: x -> y -> z.This means that x is divided by y, and y is divisible by z.That is, x = ay and y = bz (a and b - integers, a = x: y and b = y: z).Therefore, x = ay = a (bz) = abz.It follows that x divided by z (potomuchto x: z = (abz): z = abz: z = ab (z: z) = ab * 1 = ab).1. (10-11).Number 1/42 can be represented in the form of a periodic decimal fraction.The period begins with two decimal places (two) and consists of 6 numbers: 238095 (1/42 = 0.0238095238 ...).Set it is possible, for example, by dividing by 42 one "column".Since 1997, when divided by 6 gives the remainder 5, 1997, I figure recorded the same number as the fifth - zero, and the next - a nine.Hence, the new number is greater.Infinite periodic decimals are indicated as follows: 1/42 = 0.0 (238095).2. (10-11).You can.Let ABC - this equilateral triangle.Note on the AC side point D, so that AD = DC.Note on side BC point E, so that BE = CE.Note on the segment BD point F, so that BF = AF.Note on the segment AF point G, so that AG = GF.DEC (equilateral), DGF (equilateral), BED (BE = ED), BFA (BF = FA), AGD (AG = AD) - the required equilateral triangles.The possibility of constructing the conditions specified are distinct and equilateral triangles are almost obvious from the figure, a rigorous proof of these assertions is long and uninteresting and is not included here.There are other options for cutting.3. (10-11).First put on the two scales to 13 coins, then (if the balance is in equilibrium) uberm them and put on 11 more of who took no, then 9,7,5,3 and 1 to as long as one of the bowls do not outweigh .If this does not happen, then after the seventh weighing (when the scales will be only one coin) will be only one coin, which was not involved in weighing.It is false.If some kind of weighing bowl outweighed, then the false coin is in the other side.The total number of coins in this bowl denote 2k + 1 (every time we we place on one side of an odd number of coins), and we've used 7-k weighings, with each coin weighed no more than once.Therefore, it remains to find a false coin in a group of (2k + 1) ones in k weighings, each coin weighing not more than once.This can be done to break all the coins in the group, except one, divided into pairs k and the weight of the coins successively compare each pair.If some weighing balance is violated, then the lighter coin is counterfeit.Otherwise, false coin - the remaining single.

Competition in Physics

On a sheet with the task was given this explanation:After the task number in parentheses indicates how classes, this problem is recommended.To solve problems not of their class is allowed.Pay attention to the quality of physical explanation: the work, which is good, with explanations resolved two or three tasks will be highly appreciated - higher than the work in which there are fuzzy reasoning on many tasks.

Assignments1. (7-9) Compares two experiences.1. On the table is laid wooden ruler thickness of 1-2 mm and a length of 50 cm so that slightly more than half the line lay on the table, and the rest hung.If you click on the overhanging part, the line will bend easily.2. Line also is laid, but the portion of the line that is on the table is laid a sheet of paper.Now, if you quickly press the overhanging part, ruler became like heavier.Explain why this is so and describe the phenomena that occur when pressing on the line.2. (7-9) It is possible to imagine such a possibility, that in some countries the duration of the day was defined as the time between adjacent sunrise.What time of year in this country, "day" was the longest, and in some - the shortest.3. (7-9) Why eventually radius bend of the river slowly increases?4. (9-11) a) Try to estimate the order of magnitude of the coefficient of friction end of your pen on the desk or paper as follows: first put pen almost vertically, propped on her finger, and then tilt the handle up until it begins to slip.b ) Estimate the order of work on writing your work at the competition in physics.There is a view only the mechanical work on the printing on the paper;costs of other forms of energy are not counted.Note: If the task requires something to evaluate in order of magnitude, the greater accuracy is required;need just an order of magnitude when the error is 10 times is quite acceptable.5. (9-11) Contemporary Galileo, professor at the University of Padua Sanktorius invented the first thermometer, which was a glass ball filled with air and the glass tube is connected to the atmosphere.The tube is placed in a drop of water, which overlaps message cavity vessel with the external environment.With this device Sanktorius are diagnosed his patients (he was a doctor).The patient took the ball in his mouth, and a minute later on the Status of water drops Sanktorius determines what the patient fever.Depend on whether such a thermometer readings of the weather, in particular whether they will be different in clear weather and rain?6. (10-11) Try to invent a device that creates a magnified image of the object and it does not use a curved surface.7. (10-11) Try to come up with such an experience, in which the charge is moved, and the magnetic field does not arise.

In summing up all the tasks are considered equivalent (paragraphs a) and b) of the problem 4 were considered separate tasks).Reply to the competition in physics1. The thing is, of course, a piece of paper.However, the line has become "harder" is not due to the extra weight of a piece of paper (its mass is very small), and atmospheric pressure (this is known enough experience to demonstrate the atmospheric pressure).When pressing the sharp edge of the line hangs it rotates around the edge of the table by lifting the middle of the sheet of paper, whereby a vacuum is created under the sheet (air pressure below atmospheric) and pressures above the plate - atmospheric (or slightly more).As a result, a paper sheet, and, consequently, the line force directed downwards.The resulting space region with reduced pressure is communicated with the atmosphere only through the "channels" along the edges of the ruler, in which also the pressure is less than atmospheric (for two reasons: due to the vacuum existing in the fact that the air moves through "channels" on the paper a high speed).If the force due to the pressure difference in the channels and air would be enough to drag the table pressed him atmospheric pressure edges of the paper (with a strong impact on the ruler), channels slam shut, the air inlet under the sheet of paper at a time to stop and line will seem heavy.This description has been compiled on the basis of the jury's own observations.In various literature and the works of the participants of the tournament there are other explanations, slightly different than shown.All explanations, correctly reflecting the essence of the phenomenon, the jury will be counted as correct.2. It is clear that since the "Earth rotates once on its axis in 24 hours, at any time of the year" (as written almost everything, but a lot on it and stopped), the difference between adjacent sunrise will be less than 24 hours when the day is increasing and more 24 hours increased when the night.Ie from the date of the winter solstice (21 or December 22) until the day of the summer solstice (21 or June 22) night in this country will be less than 24 hours, and at other times - more.It is known that the highest intensity change of day length reaches during the spring and fall equinoxes.(You can check this on the calendar, but you can find a purely geometrical reasons - think what.) So, the shortest day in this country - the vernal equinox (March 21 or 20 depending on the year), and the longest - on the day of autumn equinox (September 23).(Recall that the tournament was held on September 28.) The exact values of the times of sunrise and sunset for Moscow can be found, for example, in "Diary of Moscow schoolchildren."Equinox(spring and autumn) - the moment of passing the center of the Sun in its apparent motion on the celestial sphere through the celestial the equator.At this time, the length of day and night is the same.Solstice(summer and winter) - moments when the height of the sun above the horizon at noon, the maximum and minimum.3. This task (question) the jury stuck in option at the last moment as a problem for younger students (as problems for the lower grades in the variant lacking), hoping to get around this answer:"Because of the centrifugal force of the water interacts strongly with the outer coast than with internal and stronger blurs it, thereby increasing the turning radius (bend) ".However, immediately after the commencement of the audit work it became clear that the issue is not as simple as it was supposed to, and clearly not for the junior.Briefly turning process River is as follows.On the right section of the river is known, the maximum speed at the center ("Bistrica") and decreases as it approaches the shore due to the friction of them.At the turn of water tends to inertia continue straight for her turn must apply a force, with the larger, the more its speed (by rotating the ball of mass m on a string in a circle of radius R with a velocity v thread tension force is mv2/ R).The result is a sort of water velocity and flow rate in the outer bank of rotation is greater.We assume the obvious, that the intensity of erosion of the coast increases with increasing flow rate.Also, we assume that a straight section of the river, this rate is an equilibrium, since there is a form of the shore does not change with time (speed increases the width of the river increases, which leads to the erosion of beaches, t. E. To increase the width of the bed. But because of the this rate of flow is reduced, leading to the shores of namyvaniyu (deposition of sedimentary rocks). And vice versa.).Then, if a straight section of the flow velocity at both banks was an equilibrium, then when you turn after the "sorting" on the outer coast of the speed is greater than the equilibrium and it will be eroded, and at the inner - less than the equilibrium, and it will pan out.In addition to "sort" the water velocity at the turn of the river there is another effect: after the "slow" the water was near the inner bank, and "fast" - from the outside, "slow" the water begins to "crush" the "fast", which leads to an even greater increase in the rate of the latter.On the other hand, the "fast" water after the interaction with the external bank becomes turbid (there are eroded from the shore rock particles) and loses its speed.Then she pushed to the inner bank of the available water flows faster.At the inner bank, in turn, loss occurs "haze" in the precipitate, i.e.additional namyvanie it.The above reasoning is a simplified description of the process at the level of the school curriculum in physics, which obviously does not take into account many important circumstances, so it is certainly not correct in terms of geology.As it turned out, this task was interested in even the Albert Einstein, in which "Reasons Education meanders in rivers and so-called Beer law "can find an explanation for the many features of the flow of rivers, for example, why the rivers of the northern hemisphere blur mainly the right bank, and the south - left.And in 1997 on this topic has been defended his doctoral dissertation in which the problem is not completely solved (so far!).4. Somehow, this problem has caused discontent decisive (maybe because we are the most essential part of their work, not mechanical, not interested): "if so try often, all the tables in the school will be inscribed."Nevertheless, many approached correctly to estimate the friction coefficient.(Although at the last moment, when the tasks have already been handed out, the jury became worried - because if you specify ktr= 0.1, in a wide range, from 1 to 0.01, the error does not exceed a given accuracy. However, students on this temptation did not fall . And rightly so - because the jury will evaluate the argument, not the final figure.)In the figure all the forces acting on the pen are indicated by a bold arrow.Rxand Ry, which are not real forces, and the projections R, shown for convenience only and therefore designated by arrows a lower fat content).The vertical component of the forces of reaction finger and paper should be equal (for example, from the condition that 0 torques about the center of the pen - or pen begins to rotate, we believe that the mass m pens distributed uniformly along its length).Together, they must balance the force of gravity mg (otherwise the handle will fail under the table), so they are both equal to mg / 2 (N = mg / 2 and Rx= mg / 2).Pen fall when the friction force F on the papertrbecomes smaller, than the horizontal pressure force component pen on paper.But where does this component, because the handle simply presses down its gravity?No, it still relies on a finger to solve this problem and the response of this support R and gives us the horizontal component of Rx.Two angle indicated in the figure symbol a (alpha) as well as angles with mutually perpendicular sides: the reaction force of the finger R, acting on pen, fountain pen perpendicular and the vertical projection of the horizontal force perpendicular to the paper surface.By definition, the tangent of an acute angle in a right triangle is equal to the ratio of the leg opposite the angle to an adjacent.Then TG = A Rx/ (mg / 2) (as defined by the tangent to the angle a, which "rests \" finger).Hence, Rx= (mg / 2) tg a.This force (Rx) can be balanced only by the force of static friction Ftr.p.pen on paper, which, in turn, can not exceed the force of friction Ftr= kN = k (mg / 2}. Therefore, at the time, when the pen in a continuous decrease in the angle of a fall,Rx= kN(mg / 2) tg a = k (mg / 2)ktr= tg aThus, estimating the angle of a tangent and finding it (and even better, you can experiment by definition, building a right-angled triangle containing the angle a, measure its legs of improvised means, such as the edge of a sheet of graph paper, and calculating their ratio), we obtain an estimate of the coefficient of friction.Now just find the mechanical work - it is the product of the frictional force on the handle passed way. It remains to estimate the weight of the handle and assess the overall length written by you (for example, assuming that all of the letters - is the height of squares corresponding to your handwriting) - and the problem is solved.

5. Actually Sanktorius are diagnosed "for 10 pulse beats," but to solve it is not essential, but only adds to the long text of the problem.Maybe because the problem could not read to the end, it was making less than we expected.Since the first thermometer was connected with the atmosphere, it influenced his work.Moreover, the thermometer affect atmospheric conditions at the time of its manufacture.Indeed, if the pressure inside the tube is increased due to increased temperature of the patient, the doctor could see that the droplet has moved by a greater amount.But if the air pressure in the day was more than daily calibration of the thermometer, the droplet could remain in place or does not reach a normal level of temperature.But of the many messages Gidromettsenrta all already known link between atmospheric pressure and clear or rainy weather.So the weather affect the reading of the thermometer.But on the other hand, do History has brought to us information about the doctor-charlatan, inaccurately diagnosed patients with fever?Rather, Sanktorius before measuring the temperature of the patient, the thermometer calibration is performed by measuring the temperature at, and only then - in a patient.Jury these details are not known for certain, but indirect evidence of them is the fact that the patient briefly ("for 10 pulse beats") held a thermometer at home, because he was already heated professor!This example shows that it may not be the right answer monosyllabic (effect-no effect), is important to correctly reasoning that, depending on the additional assumptions lead to different (but good!) conclusions.6. A list of devices that satisfy the conditions of the problem, these students: the camera obscura, crossed prisms, holography (hologram recorded at a single wavelength and reproduced at the other), the diffraction grating with concentric rings, different ways to create optical density inhomogeneities in the medium (lighter in the air is dissolved in water, salt crystals).There are other options, but they are not offered to the students and are very hard to describe, so no more description.7. Take the spherical surface (e.g., a balloon) at which electric charge is uniformly distributed.To begin to change its radius without moving her center.Located on the surface charges will obviously move, with the process of moving is centrally symmetric.Consequently, all of the physical phenomena that result from this process should also have a center of symmetry. *The only possible (from geometrical considerations) in this situation, the magnetic field configuration is this: the magnetic field lines diverge uniformly in all directions from the center of the sphere.But this is impossible, sinceMagnetic lines of force may not be endless. *Obviously, the condition of the problem satisfy any centrally symmetric motion of the charges and the superposition of such movements.The jury is not known whether such a situation in the other cases.Note:For an asterisk claims in the scientific world there is no single opinion.We rely on the school curriculum and are not going to contribute to the debate on these issues in modern physicsInformation about the competition jury in PhysicsOn estimates of the competition in physics tournament them.Lomonosov 1997Competitions in physics and mathematics tournament it.Lomonosov Moscow Olympiad in mathematics, Tournament cities and some other similar events are held the same korrellirovanoe a lot of people, so in all these events is used for many years has developed a system of ratings:0 - -.- + + / 2 + - +.+!0, the problem is not recorded in the- In the solution contained only wrong, have no relation to the correct solution or contained in the problem statements-.the problem is not solved, but the decision has at least one nontrivial true statement pertaining to the right decision and not contained in (all estimates are reported to participants, so stamped assessment "minus" in this case may give the student a misconception that in his work incorrectly absolutely everything, including this very statement)- + Problem is solved correctly, but the decision has significant progress in the right direction.For the right decision, but their absence, usually put + - + / 2 in the problem can not be solved sorting options - the correct answer without explanation.In the task, where necessary, for example, to prove the necessity and sufficiency of something - for the right solution to only one of the parts.In other problems - if the solution has a sufficiently large number of true moves, but to build one right solution still can not.+ - The right decision with a significant failure.+.the problem is solved correctly, but there are insignificant flaws+ Problem is solved correctly+!the problem is solved correctly and in the decision contains the original and / or substantially beyond the school curriculum (corresponding class) points that the jury considers it necessary to emphasize.Spelling errors are not taken into account (because it is a competition in physics, not in Russian, but also because the primary school children in solving complex problems of the Contest can not do without grammatical constructions that they have not learned in school).Although, unfortunately, there are so many errors ...Arithmetic errors do not affect the meaning of the solution, not taken into account (is reasonable to assume that students 7th grade and older are able to perform arithmetic operations, such errors are not caused by lack of knowledge, and other factors).If the work has only correct strikeout decision, it is counted as correct.If the solution of the problem can not be obtained by brute force of a small number of options (yes / no increase / no change / decrease), and her work has several options, including the right, count the correct option (even if it is crossed out).Available for solving incorrect statements that have no semantic connections with the rest of the decision, the assessment is not reduced.The criteria for determining the winnersGrade 6 and underDiploma if one + / 2 or betterOfficially, the tournament is held for students grades 7-11, but talented schoolchildren also successfully participate in the tournament.This year, the students of class 5 and younger participate in the competition in physics is not taken.The competition was attended by two sixth-grader:Gaifullin Sergey school '10 Zhukovsky, Moscow Region,evaluation - +.+.0 0 0 0 0(This is a very good result for grade 6)Kolychev Alexander School '58 Moscowevaluation 0 + / 2 + 0/2 0 0 0 0(The discussion of the jury also decided to award him a diploma, that is the criterion for grade 6 is purely formal)7 and 8 classesAround if one + / 2 or betterAround if the three - + or betterDiploma if one + - or betterGrade 9 and olderAround if one + - or betterAround, if three - + or betterdiploma if the two + - or betterThe phrase "or better" does not mean subjective view "better or worse", and the presence in the specified number or those estimates or higher.Any formal justification for the above criteria for determining the winners do not exist.The decision was made after checking all the works under discussion based on the experience of last year's results, the actual situation at work, personal opinions of the members of the jury.CheckThe audit involved 7 (NN Konstantinov, undergraduate and graduate students, physics).Each work independently verified first 2 times, then recheck again (usually the third person), which, when there is a discrepancy between the estimates and the estimates and its own opinion to hold discussions with the other reviewers (sometimes quite tense), and set the final evaluation.At the closing ceremony the students could challenge them (this year, wishing there was in the past - were).When checking the work of the different classes were presented the same requirements for the physical side solutions, but takes into account different levels of intellectual development of students of different ages in the evaluation of logical correctness and validity of the decision.The difference between the classes into account in the criteria for determining the winners.Of course, when checking have been developed certain formal criteria, but each work still tested individually, in accordance with common sense, and the criteria themselves are often formulated in the form of words or phrases such as "prism", "centrifugal force".Bringing them to the formal compliance with all the assessments require breaching a lot of work for what the jury had no opportunity now (see. The last paragraph), and the publication of the criteria without such a reduction can cause massive resentment among the tournament participants, their teachers and parents.Therefore, the criteria are not published.7 checks - to put it mildly, is not enough (in mathematics at about the same amount of work they were 37).But more than just was not wishing, most physicists were limited excuses and valuable comments and innovations.If any of the readers willing to make a personal contribution to the work tasks Tournament next year, its implementation and verification work, the jury happy to support this initiative.Selection problemsAt the competition were invited to a lot of problems (7, one of them with two points), but to get high score was enough to solve a small part of them, as can be seen from the above-stated criteria.For each task, indicate which classes students recommended to solve it (for each class - at least 3 tasks to provide student the opportunity to choose a favorite subject for the task and to avoid a situation where the student can not solve any problem, because I do not know exactly the formulas and definitions that are needed for this).Students can solve problems and not the class, summarizing the results of all the tasks are taken into account equally.This rule came as follows: The tournament has been held for more than 20 years and its rules have changed over time.In particular, in 1992/93, and a few years before the tournament formally held for grades 7-9, and for 10 and 11 classes at the same time and in the same places Tournament held cities in mathematics.But the ubiquitous high school students still leaked on Turlov.Combat it was useless and in the 1993/94 school year, they were invited to the event "officially", and for them was made a separate option in physics.Due to the fact that it successfully solved and several (about 15) schoolchildren in HS 1995/96the jury decided to conduct an experiment and offer a "senior" mladsheklassnikov all, it was found that almost half of mladsheklassnyh degrees in physics (mainly in part) due to problem solving "older" version.It became apparent that their division would be extremely unfair to such students.The complexity of the task (for the class) approximately corresponds to the level of regional Olympiads (Juniors - fans challenges - can solve the problems of high school, and high school students, especially for fans of a problem is the university level ( 7, about the movement of the charge and the magnetic field). Relatively low complexity of most tasks and assessment criteria are due to lack of time (7 competitions for 5 hours) and low (unfortunately) the general level of school physical education in Moscow (much lower than the mathematical or humanitarian).The content of the competition problems in physics this tournament is somewhat different from those of the school, district, city and university competitions.Jury tries to avoid formal synthetic problems, a key element of the decision which is the formal knowledge and combination formulas and definitions (such problems and so much on the above activities and entrance examinations), mathematical problems "in the physical Skin" (in the tournament have a competition in mathematics) very simple (comforting) problems (students should not be in vain to spend their decision time and effort required in other competitions).Preference is given to quality problems and issues, a simple experimental tasks (which are shown or in the audience, and each participant can do them in place), the tasks for which it is necessary to understand the content of the physical process or phenomenon, evaluation tasks.The jury also fundamentally waived task, which is difficult to check.Back to the list of contests

Competition in ChemistryTasks for competition in chemistry providedby the Moscow Chemical LyceumHouse of scientific and technical creativity of youth in Moscow.Solving problems of the above-mentioned organization, unfortunately, are not provided for that tournament organizing committee apologizes.Participants of the tournament that the jury expects to receive from them three of the proposed solutions of problems of choice.For each task to grade on a scale from 0 to 5. The final evaluation is the sum of scores for the three tasks otsennnye highest number of points than the other (points a, b, c, objectives 5 assessed on a 5-point scale individually and summarizing the results were considered separate tasks; unfortunately, it has not been announced to participants in advance).If such problems but three student solved yet sufficient other, the amount increased by 1.Tasks1. (Grades 9-11) According to the information preserved in folk tradition, in the first century of our millennium there was the idea that there is metal as much and planets.Write the chemical symbols of these metals.2. (Grade 9-10) As, using as the starting materials iron filings, salt, baking soda, chalk acid battery for refueling, water and air, to obtain various iron compounds.The equations of the corresponding reactions, indicating the conditions of their implementation.3. (9-10 grade) proposed a formula of inorganic substances A and for which you have the following transformation:A + B -> BDT-> B + LE + F -> WA + F -> AndG + H -> R + D+ E B -> DB + D -> DA + H -> R + EB +