Essential Amino AcidsAmino acidsare organic compounds which contain both an amino group and a carboxyl group. According to Tillery, et al., the human body can synthesize all of the amino acids necessary to build proteins except for the ten called the "essential amino acids", indicated by asterisks in theamino acid illustrations. An adequate diet must contain these essential amino acids. Typically, they are supplied by meat and dairy products, but if those are not consumed, some care must be applied to ensuring an adequate supply. They can be supplied by a combination of cereal grains (wheat, corn, rice, etc.) and legumes (beans,peanuts, etc.). Tillery points out that a number of popular ethnic foods involve such a combination, so that in a single dish, one might hope to get the ten essential amino acids. Mexican corn and beans, Japanese rice and soybeans, and Cajun red beans and rice are examples of such fortuitous combinations.

Half-life(t12) is the amount of time required for a quantity to fall to half its value as measured at the beginning of the time period. While the term "half-life" can be used to describe any quantity which follows anexponential decay, it is most often used within the context ofnuclear physicsandnuclear chemistrythat is, the time required, probabilistically, for half of the unstable, radioactiveatomsin a sample to undergoradioactive decay.The original term, dating toErnest Rutherford's discovery of the principle in 1907, was "half-life period", which was shortened to "half-life" in the early 1950s.[1]Rutherford applied the principle of a radioactiveelement'shalf-life to studies of age determination ofrocksby measuring the decay period ofradiumtolead-206.Half-life is used to describe a quantity undergoing exponential decay, and is constant over the lifetime of the decaying quantity. It is acharacteristic unitfor the exponential decay equation. The term "half-life" may generically be used to refer to any period of time in which a quantity falls by half, even if the decay is not exponential. The table on the right shows the reduction of a quantity in terms of the number of half-lives elapsed. For a general introduction and description of exponential decay, seeexponential decay. For a general introduction and description of non-exponential decay, seerate law. The converse of half-life isdoubling time.Contents[hide] 1Probabilistic nature of half-life 2Formulas for half-life in exponential decay 2.1Decay by two or more processes 2.2Examples 3Half-life in non-exponential decay 4Half-life in biology and pharmacology 5See also 6References 7External linksProbabilistic nature of half-life[edit]

Simulation of many identical atoms undergoing radioactive decay, starting with either 4 atoms per box (left) or 400 (right). The number at the top is how many half-lives have elapsed. Note the consequence of thelaw of large numbers: with more atoms, the overall decay is more regular and more predictable.A half-life usually describes the decay of discrete entities, such as radioactive atoms. In that case, it does not work to use the definition "half-life is the time required for exactly half of the entities to decay". For example, if there are 3 radioactive atoms with a half-life of one second, there will not be "1.5 atoms" left after one second.Instead, the half-life is defined in terms ofprobability: "Half-life is the time required for exactly half of the entities to decayon average". In other words, theprobabilityof a radioactive atom decaying within its half-life is 50%.For example, the image on the right is a simulation of many identical atoms undergoing radioactive decay. Note that after one half-life there are notexactlyone-half of the atoms remaining, onlyapproximately, because of the random variation in the process. Nevertheless, when there are many identical atoms decaying (right boxes), thelaw of large numberssuggests that it is avery good approximationto say that half of the atoms remain after one half-life.There are various simple exercises that demonstrate probabilistic decay, for example involving flipping coins or running a statisticalcomputer program.[2][3][4]Formulas for half-life in exponential decay[edit]Main article:Exponential decayAn exponential decay process can be described by any of the following three equivalent formulas:

where N0is the initial quantity of the substance that will decay (this quantity may be measured in grams, moles, number of atoms, etc.), N(t) is the quantity that still remains and has not yet decayed after a timet, t12is the half-life of the decaying quantity, is apositive numbercalled themean lifetimeof the decaying quantity, is a positive number called thedecay constantof the decaying quantity.The three parameters,, andare all directly related in the following way:

where ln(2) is thenatural logarithmof 2 (approximately 0.693).[show]Click "show" to see a detailed derivation of the relationship between half-life, decay time, and decay constant.

By plugging in and manipulating these relationships, we get all of the following equivalent descriptions of exponential decay, in terms of the half-life:

Regardless of how it's written, we can plug into the formula to get as expected (this is the definition of "initial quantity") as expected (this is the definition of half-life) ; i.e., amount approaches zero astapproaches infinityas expected (the longer we wait, the less remains).Decay by two or more processes[edit]Some quantities decay by two exponential-decay processes simultaneously. In this case, the actual half-life T12can be related to the half-lives t1and t2that the quantity would have if each of the decay processes acted in isolation:

For three or more processes, the analogous formula is:

For a proof of these formulas, seeExponential decay#Decay by two or more processes.Examples[edit]

Half life demonstrated using dice in aclassroom experimentFurther information:Exponential decay Applications and examplesThere is a half-life describing any exponential-decay process. For example: The current flowing through anRC circuitorRL circuitdecays with a half-life ofor, respectively. For this example, the termhalf timemight be used instead of "half life", but they mean the same thing. In a first-orderchemical reaction, the half-life of the reactant is, where is thereaction rate constant. Inradioactive decay, the half-life is the length of time after which there is a 50% chance that an atom will have undergonenucleardecay. It varies depending on the atom type andisotope, and is usually determined experimentally. SeeList of nuclides.The half life of a species is the time it takes for the concentration of the substance to fall to half of its initial value.Half-life in non-exponential decay[edit]Main article:Rate equationThe decay of many physical quantities is not exponentialfor example, the evaporation of water from a puddle, or (often) the chemical reaction of a molecule. In such cases, the half-life is defined the same way as before: as the time elapsed before half of the original quantity has decayed. However, unlike in an exponential decay, the half-life depends on the initial quantity, and the prospective half-life will change over time as the quantity decays.As an example, the radioactive decay ofcarbon-14is exponential with a half-life of 5730 years. A quantity of carbon-14 will decay to half of its original amount (on average) after 5730 years, regardless of how big or small the original quantity was. After another 5730 years, one-quarter of the original will remain. On the other hand, the time it will take a puddle to half-evaporate depends on how deep the puddle is. Perhaps a puddle of a certain size will evaporate down to half its original volume in one day. But on the second day, there is no reason to expect that one-quarter of the puddle will remain; in fact, it will probably be much less than that. This is an example where the half-life reduces as time goes on. (In other non-exponential decays, it can increase instead.)The decay of a mixture of two or more materials which each decay exponentially, but with different half-lives, is not exponential. Mathematically, the sum of two exponential functions is not a single exponential function. A common example of such a situation is the waste of nuclear power stations, which is a mix of substances with vastly different half-lives. Consider a sample containing a rapidly decaying element A, with a half-life of 1 second, and a slowly decaying element B, with a half-life of one year. After a few seconds, almost all atoms of the element A have decayed after repeated halving of the initial total number of atoms; but very few of the atoms of element B will have decayed yet as only a tiny fraction of a half-life has elapsed. Thus, the mixture taken as a whole does not decay by halves.

Solubility Rules!A basic knowledge of which compounds are soluble in aqueous solutions is essential for predicting whether a given reaction might involve formation of a precipitate (an insoluble compound).The following guidelines are generalizations. A substance is classified as insoluble if it precipitates when equal volumes of 0.1 M solutions of its components are mixed. Keep in mind, however, that no substance is completely insoluble. Substances listed as insoluble are, at some level, partially soluble. The magnitude of the ion product constant (Ksp) for the appropriate solubility equilibrium should be examined. LargerKspvalues indicate greater solubility; smallerKspvalues indicate lesser solubility.The symbol "" is used here to signify the 'double-arrrow' symbol for a chemical equilibrium. The symbol "=>" is used here to signify the 100% dissociation of a compound into its electrolyte ions in aqueous solution. The subscript "(s)" following a species indicates that it is a solid. The subscript "(aq.)" following a species indicates that it is in aqueous solution.Rule 1.All compounds of Group IA elements (the alkali metals) aresoluble.For example, NaNO3, KCl, and LiOH are all soluble compounds. This means that an aqueous solution of KCl really contains the predominant species K+and Cl-and, because KCl is soluble, no KCl is present as a solid compound in aqueous solution:KCl(s)=> K+(aq.)+ Cl-(aq.)Rule 2.All ammonium salts (salts of NH4+) aresoluble.For example, NH4OH is a soluble compound. Molecules of NH4OH completely dissociate to give ions of NH4+and OH-in aqueous solution.Rule 3.All nitrate (NO3-), chlorate (ClO3-), perchlorate (ClO4-), and acetate (CH3COO-or C2H3O2-, sometimes abbreviated as Oac-) salts aresoluble.For example, KNO3would be classified as completely soluble by rules 1 and 3. Thus, KNO3could be expected to dissociate completely in aqueous solution into K+and NO3-ions: KNO3=> K+(aq.) + NO3-(aq.)Rule 4.All chloride (Cl-), bromide (Br-), and iodide (I-) salts aresolubleexcept for those of Ag+, Pb2+, and Hg22+.For example, AgCl is a classic insoluble chloride salt:AgCl(s) Ag+(aq.)+ Cl-(aq.)(Ksp= 1.8 x 10-10).Rule 5.All sulfate ( SO4=) compounds aresolubleexcept those of Ba2+, Sr2+, Ca2+, Pb2+, Hg22+, and Hg2+, Ca2+and Ag+sulfates are only moderately soluble.For example, BaSO4is insoluble (only soluble to a very small extent):BaSO4(s) Ba2+(aq.)+ SO42-(aq.)(Ksp= 1.1 x 10-10).Na2SO4is completely soluble:Na2SO4(s)=> 2 Na+(aq.)+ SO42-(aq.).Rule 6.All hydroxide (OH-) compounds areinsolubleexcept those of Group I-A (alkali metals) and Ba2+, Ca2+, and Sr2+.For example, Mg(OH)2is insoluble (Ksp= 7.1 x 10-12).NaOH and Ba(OH)2are soluble, completely dissociating in aqueous solution:NaOH(s)=> Na+(aq.)+ OH-(aq.), a strong baseBa(OH)2(s)=> Ba2+(aq.)+ 2OH-(aq.)(Ksp= 3 x 10-4)Rule 7.All sulfide (S2-) compounds areinsolubleexcept those of Groups I-A and II-A (alkali metals and alkali earths).For example, Na2S(s) 2Na+(aq.)+ S2-(aq.)MnS is insoluble (Ksp= 3 x 10-11).Rule 8.All sulfites (SO3=), carbonates (CO3=), chromates (CrO4=), and phosphates (PO43-) areinsolubleexcept for those of NH4+and Group I-A (alkali metals)(see rules 1 and 2).For example, calcite, CaCO3(s)Ca2+(aq.)+ CO3=(aq.)(Ksp= 4.5 x 10-9).

D and L Are Outmoded and WrongStudents who take biochemistry are exposed to an old, confusing, and often incorrect method of specifying configurations at chiral centers as D or L. This scheme was devised in the early years of this century by Emil Fischer, a German organic chemist who worked extensively with carbohydrates. Here's how it is supposed to work.The compound glyceraldehyde, HOCH2CH(OH)CHO, was chosen as the standard for defining configuration. The enantiomer that rotates plane polarized light clockwise (+) was arbitrarily labeled D The other enantiomer (-) became L.

As shown, the assignments in modern notation are R and S, respectively. (Note: it willnotalways work out that D = R and L=S; this is an accident here.) The source of the D and L labels was the Latin wordsdexter(on the right) andlaevus(on the left) R comes fromrectus(right-handed) and S fromsinister(left-handed)Any other molecule containing a single chiral center was to be assigned as D or L by imagining a resemblance between the ligands on its chiral center and those in glyceraldehyde The enantiomer having the "same or similar" groups in the same places as D-glyceraldehyde becomes D. Thus, for example, the naturally occuring form of the amino acid cysteine was labeled L.

When more than one chiral center is present, similarity is defined only by the arrangement of ligands on the highest numbered chiral center, and the assignment of D or L is made on the basis of that center only. The configuration at the other centers usually is specified by giving the diastereomeric molecules entirely different names. The student is expected simply to memorize which arrangement of ligands goes with which names. For example, look at the two aldotetroses below:

The D and L were assigned on the basis of the arrangement at the lower of the two chiral centers. That the two sugars are diastereomers is specified by giving them different names.Clearly, this system is impossible to apply widely, requires extensive memorizing of structures, and is also seriously ambiguous.Consider for example, the reaction outline shown below. Here D-glyceraldehyde is converted into lactic acid by reactions thatdo not break any bonds to the chiral center!(These conversions actually have been carried out in the laboratory.)

Which structure is D-lactic acid?

The Photoelectric EffectWhat's the photoelectric effect?

It's been determined experimentally that when light shines on a metal surface, the surface emits electrons. For example, you can start a current in a circuit just by shining a light on a metal plate. Why do you think this happens?

Well...we were saying earlier that light is made up of electromagnetic waves, and that the waves carry energy. So if a wave of light hit an electron in one of the atoms in the metal, it might transfer enough energy to knock the electron out of its atom.

Okay. Now, if light is indeed composed of waves, as you suggest...What do you mean, "iflight is composed of waves"? Is there another option?

Historically, light has sometimes been viewed as aparticlerather than a wave; Newton, for example, thought of light this way. The particle view was pretty much discredited with Young'sdouble slit experiment, which made things look as though light had to be a wave. But in the early 20th century, some physicists--Einstein, for one--began to examine the particle view of light again. Einstein noted that careful experiments involving the photoelectric effect could show whether light consists of particles or waves.

How? It seems to me that the photoelectric effect would still occur no matter which view is correct. Either way, the light would carry energy, so it would be able to knock electrons around.

Yes, you're right--but thedetailsof the photoelectric effect come out differently depending on whether light consists of particles or waves. If it's waves, the energy contained in one of those waves should depend only on its amplitude--that is, on the intensity of the light. Other factors, like the frequency, should make no difference. So, for example, red light and ultraviolet light of the same intensity should knock out the same number of electrons, and the maximum kinetic energy of both sets of electrons should also be the same. Decrease the intensity, and you should get fewer electrons, flying out more slowly; if the light is too faint, you shouldn't get any electrons at all, no matter what frequency you're using.

That sounds reasonable enough to me. How would the effect change if you assume that light is made of particles?

I should give you some background information on this, first. It all began with some work on radiation by Max Planck...Planck's Constant and the Energy of a PhotonIn 1900, Max Planck was working on the problem of how the radiation an object emits is related to its temperature. He came up with a formula that agreed very closely with experimental data, but the formula only made sense if he assumed that the energy of a vibrating molecule wasquantized--that is, it could only take on certain values. The energy would have to be proportional to the frequency of vibration, and it seemed to come in little "chunks" of the frequency multiplied by a certain constant. This constant came to be known asPlanck's constant, orh, and it has the value

This doesn't make any sense to me. I think I'llgo askDr. Mahan whatJmeans.

That's a pretty small constant.

Yes, but it was an extremely radical idea to suggest that energy could only come in discrete lumps, even if the lumps were very small. Planck actually didn't realize how revolutionary his work was at the time; he thought he was just fudging the math to come up with the "right answer," and was convinced that someone else would come up with a better explanation for his formula.

I guess Einstein must have taken him seriously, though.

Quite seriously. Based on Planck's work, Einstein proposed that light also delivers its energy in chunks; light would then consist of little particles, orquanta, calledphotons, each with an energy of Planck's constant times its frequency.

In that case, the frequency of the lightwouldmake a difference in the photoelectric effect.

Exactly. Higher-frequency photons have more energy, so they should make the electrons come flying out faster; thus, switching to light with the same intensity but a higher frequency should increase the maximum kinetic energy of the emitted electrons. If you leave the frequency the same but crank up the intensity,moreelectrons should come out (because there are more photons to hit them), but they won't come out anyfaster, because each individual photon still has the same energy.

And if the frequency is low enough, then none of the photons will have enough energy to knock an electron out of an atom. So if you use really low-frequency light, you shouldn't getanyelectrons, no matter how high the intensity is. Whereas if you use a high frequency, you should still knock out some electrons even if the intensity is very low.

Quite right. Therefore, with a few simple measurements, the photoelectric effect would seem to be able to tell us whether light is in fact made up of particles or waves.

So did someone do the experiment? Which way did it turn out?

In 1913-1914, R.A. Millikan did a series of extremely careful experiments involving the photoelectric effect. He found that all of his results agreed exactly with Einstein's predictions about photons, not with the wave theory. Einstein actually won the Nobel Prize for his work on the photoelectric effect, not for his more famous theory of relativity.

Then lightismade of particles! But wait...what about the two-slit experiment? I don't see how light could make an interference pattern like that, unless it was made of waves.

Yes, I'm afraid it's a bit more complicated than that. Some experimental results, like this one, seem to prove beyond all possible doubt that light consists of particles; others insist, just as irrefutably, that it's waves. We can only conclude that light is somehow both a wave and a particle--or that it's something else we can't quite visualize, which appears to us as one or the other depending on how we look at it.

Photoelectric effectFrom Wikipedia, the free encyclopediaPhotoelectric effect

Adiagramthat shows howelectronsareemittedfrom a metal plate

Light-matter interaction

Low energy phenomenaPhotoelectric effect

Mid-energy phenomenaCompton scattering

High energy phenomenaPair production

Thephotoelectric effectis aphenomenoninphysics. The effect is based on the idea that electromagnetic radiation is made of a series of particles calledphotons.[1]When a photon hits anelectronon a metal surface, the electron can beemitted.[2]The emitted electrons are calledphotoelectrons.[1]The effect is also called theHertz Effect,[3][4]because it was discovered byHeinrich Rudolf Hertz, but this name is not used often.The photoelectric effect has helpedphysicistsunderstand thequantumnature oflightandelectrons. The concept ofwaveparticle dualitywas developed because of the photoelectric effect.[2]Mechanism[change|change source]Not every electromagnetic wave will cause the photoelectric effect, only radiation of a certain frequency or higher will cause the effect. The minimum frequency needed is called thecutoff frequencyor threshold frequency. The cutoff frequency is used to find thework function, w, which is the amount of energy holding the electron to the metal surface.[1]The work function is a property of the metal and is not affected by the incoming radiation.[1]If a frequency of light strikes the metal surface that is greater than the cutoff frequency, then the emitted electron will have some kinetic energy.The energy of a photon causing the photoelectric effect is found through E = hf = KE + w, where h is Planck's constant, 6.626X10^(-34) J*s, f is the frequency of the electromagnetic wave, KE is the kinetic energy of the photoelectron and w is the work function for the metal.[1]If the photon has a lot of energy,compton scattering(~thousands of eV) orpair production(~millions of eV) may take place.The intensity of the light does not cause ejection of electrons, only light of the cut off frequency or higher can do that. However increasing the intensity of light will increase the number of electrons being emitted, so long as the frequency is above the cut off frequency.[1]History[change|change source]The first observation of the photoelectric effect was by Heinrich Hertz in 1887. He reported that a spark jumped more readily between two charged spheres if light was shining on them.[1]Further studies were done to learn about the effect observed by Hertz, however it wasn't until 1905 that a theory was proposed that explained the effect completely. The theory was proposed by Einstein and it made the claim that electromagnetic radiation had to be thought of as a series of particles, called photons, which collide with the electrons on the surface and emit them.[1]This theory ran contrary to the belief that electromagnetic radiation was a wave and thus it was not recognised as correct until 1916 when Robert Millikan performed a series of experiments using a vacuum photo-tube to confirm the theory.[1]

CAcT HomePageThe Ideal Gas LawSkills to develop Explain all the quantities involved in the ideal gas law. Evaluate the gas constant R from experimental results. CalculateT, V, P,or n of the ideal gas law,P V = n R T. Describe the ideal gas law using graphics.The Ideal Gas LawThe volume (V) occupied bynmoles of any gas has a pressure (P) at temperature (T) in Kelvin. The relationship for these variables,P V=n R T,whereRis known as thegas constant, is called theideal gas laworequation of state. Properties of the gaseous state predicted by the ideal gas law are within 5% for gases under ordinary conditions. In other words, given a set of conditions, we can predict or calculate the properties of a gas to be within 5% by applying the ideal gas law. How to apply such a law for a given set of conditions is the focus of general chemistry.At a temperature much higher than the critical temperature and at low pressures, however, the ideal gas law is a very good model for gas behavior. When dealing with gases at low temperature and at high pressure, correction has to be made in order to calculate the properties of a gas in industrial and technological applications. One of the common corrections made to the ideal gas law is thevan der Waal's equation, but there are also other methods dealing with the deviation of gas from ideality.The Gas ConstantRRepeated experiments show that at standard temperature (273 K) and pressure (1 atm or 101325 N/m2), one mole (n= 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate thegas constantR, P V 1 atm 22.4 LR = --- = ------------ n T 1 mol 273 K

= 0.08205 L atm / (molK)When SI units are desirable,P= 101325 N/m2(Pa for pascal) instead of 1 atm. The volume is 0.0224 m3. The numberical value and units forRis 101325 N/m2 0.0224 m3R = ---------------------- 1 mol 273 K

= 8.314 J / (molK)Note that 1 L atm = 0.001 m3x 101325 N / m2= 101.325 J (or N m). Since energy can be expressed in many units, other numerical values and units forRare frequently in use.For your information, the gas constant can be expressed in the following values and units.R = 0.08205 L atm / molK Notes: = 8.3145 L kPa / molK 1 atm = 101.32 kPa = 8.3145 J / molK 1 J = 1 L kPa = 1.987 cal / molK 1 cal = 4.182 J = 62.364 L torr/ molK 1 atm = 760 torrThe gas constantRis such a universal constant for all gases that its values are usually listed in the "Physical Constants" of textbooks and handbooks. It is also listed inConstantsof our HandbookMenu at the left bottom. Although we try to use SI units all the time, the use of atm for pressure is still common. Thus, we often use R = 8.314 J / (molK) or 8.3145 J / molK.The volume occupied by one mole,n= 1, of substance is called themolar volume,Vmolar=V / n. Using the molar volume notation, the ideal gas law is:P Vmolar=R TApplications of the Ideal Gas LawThe ideal gas law has four parameters and a constant,R,P V=n R T,and it can be rearranged to give an expression for each ofP, V, norT. For example,P=n R T / V, (Boyles law)P= (n R / V)T (Charles law)These equations are Boyles law and Charles law respectively. Similar expressions can be derived forV, nandTin terms of other variables. Thus, there are many applications. However, you must make sure that you use the proper numerical value for the gas constantRaccording to the units you have for the parameters.Furthermore,n / Vis number of moles per unit volume, and this quantity has the same units as the concentration (C). Thus, the concentration is a function of pressure and temperature,C=P/R T.At 1.0 atm pressure and room temperature of 298 K, the concentration of an ideal gas is 0.041 mol/L.The Avogadros law can be further applied to correlate gas densityd(weight per unit volume orn M / V) and molecular massMof a gas. The following equation is easily derived from the ideal gas law: n M P M = --- R T V Thus, we haveP M = d R T / Md = n M / V- definition, andd = P M / R TM = d R T / PExample 1An air sample containing only nitrogen and oxygen gases has a density of 1.3393 g / L at STP. Find the weight and mole percentages of nitrogen and oxygen in the sample.SolutionFrom the densityd, we can evaluate an average molecular weight (also called molar mass).P M = d R TM = 22.4 * d = 22.4 L/mol * 1.3393 g/L = 30.0 g / molAssume that we have 1.0 mol of gas, andxmol of which is nitrogen, then (1 -x) is the amount of oxygen. The average molar mass is the mole weighted average, and thus,28.0x+ 32.0 (1 -x) = 30.0- 4x= - 2x= 0.50 mol of N2, and 1.0 - 0.50 = 0.50 mol O2Now, to find the weight percentage, find the amounts of nitrogen and oxygen in 1.0 mol (30.0 g) of the mixture.Mass of 0.5 mol nitrogen = 0.5 * 28.0 = 14.0 gMass of 0.5 mol oxygen = 0.5 * 32.0 = 16.0 gPercentage of nitrogen = 100 * 14.0 / 30.0 = 46.7 % Percentage of oxygen = 100 * 16.0 / 30.0 = 100 - 46.7 = 53.3 %DiscussionWe can find the density of pure nitrogen and oxygen first and evaluate the fraction from the density.dof N2= 28.0 / 22.4 = 1.2500 g/Ldof O2= 32.0 / 22.4 = 1.4286 g/L1.2500x+ 1.4286 (1 -x) = 1.3393Solving forxgivesx= 0.50 (same result as above)ExerciseNow, repeat the calculations for a mixture whose density is 1.400 g/L.Example 2What is the density of acetone, C3H6O, vapor at 1.0 atm and 400 K?SolutionThe molar mass of acetone = 3*12.0 + 6*1.0 + 16.0 = 58.0. Thus,d = P M / R T = (1.0 * 58.0 atm g/mol) / (0.08205 L atm / (mol K) * 400 K) = 1.767 g / LExerciseThe density of acetone is 1.767 g/L, calculate its molar mass.Confidence Building Questions Top of FormWhat is the variablenstand for in the ideal gas law,P V=n R T?

Skill:Describe the ideal gas law.Bottom of Form Top of FormA closed system means no energy or mass flow into or out of a system. In a closed system, how many independent variables are there amongn, T, VandPfor a gas?Note: an independent variable can be of any arbitrary values.Skill:The ideal gas equation shows the interdependence of the variables. Only one of them can be varied independently.Bottom of Form Top of FormWhat is the molar volume of an ideal gas at 2 atm and 1000 K?Skill:Evaluate molar volume at any condition.Bottom of Form Top of FormA certain amount of a gas is enclosed in a container of fixed volume. If you let heat (energy) flow into it, what will increase?(In a multiple choice, you may have volume, pressure, temperature, and any combination of these to choose from.)Skill:Explain a closed system and apply ideal gas law.Bottom of Form Top of FormFor a certain amount (n= constant) of gas in a closed system, how does volumeVvary with the temperature? In the following,kis a constant depending onnandP.a.V = k Tb.V = k / Tc.T V = kd.V = k T2e.V = k

Skill:Explain Charles law.Bottom of Form Top of FormBoyles law isP V= constant. A sketch ofPvsVon graph paper is similar to a sketch of the equationx y= 5. What curve(s) does this equation represent?a. a parabolab. an ellipsec. a hyperbolad. a pair of hyperbolae. a straight linef. a surface

Skill:Apply the skills acquired in Math courses to chemical problem solving.Bottom of Form Top of FormFor a certain amount of gas in a closed system, which one of the following equation is valid? Subscripts 1 and 2 refer to specific conditions 1 and 2 respectively.a.P1V1T1=P2V2T2b.P1V1T2=P2V2T1c.P1V2T1=P2V1T2d.P2V1T1=P1V2T2e.P1V2/T1=P2V1/T2

Skill:Rearrange a mathematical equation.Bottom of Form Top of FormThe gas constantRis 8.314 J / molK. Convert the numerical value ofRso that its units are cal / (molK). A unit conversion table will tell you that 1 cal = 4.184 J. Make sure you know where to find it. During the exam, the conversion factor is given, but you should know how to use it.Skill:Use conversion factors, for example: 1 cal8.314 J ------- = ? cal 4.184 J Bottom of Form Top of FormAt standard temperature and pressure, how many moles of H2are contained in a 1.0 L container?Discussion:There are many methods for calculating this value.Bottom of Form Top of FormAt standard temperature and pressure, how many grams of CO2is contained in a 3.0 L container? Molar mass of CO2= 44.One method:It containsn= 1 atm * 3 L / (0.08205 L atm / (molK) * 273 K)Bottom of Form Top of FormWhat is the pressure if 1 mole of N2occupy 1 L of volume at 1000 K?Discussion:Depending on the numerical value and units ofRyou use, you will get the pressure in various units.At 1000 K, some of the N2molecules may dissociate. If that is true, the pressure will be higher!Bottom of Form Top of FormWhat is the temperature if 1 mole of N2occupy 100 L of volume has a pressure of 20 Pa (1 Pa = 1 Nm-2)?Discussion:AtT= 240 K, ideal gas law may not apply to CO2, because this gas liquifies at rather high temperature. The ideal gas law is still good for N2, H2, O2etc, because these gases liquify at much lower temperature.Bottom of Form