The John Fisher School Summer revision calculations for AS chemists September 2016 Q1.Dilute nitric acid reacts with potassium hydroxide solution. The equation for the reaction is: HNO 3 + KOH KNO 3 + H 2 O A student investigated the temperature change in this reaction. This is the method the student used. Step 1 Put 25 cm 3 of dilute nitric acid in a polystyrene cup. Step 2 Use a thermometer to measure the temperature of the dilute nitric acid. Step 3 Use a burette to add 4 cm 3 of potassium hydroxide solution to the dilute nitric acid and stir the mixture. Step 4 Use a thermometer to measure the highest temperature of the mixture. Step 5 Repeat steps 3 and 4 until 40 cm 3 of potassium hydroxide solution have been added. The dilute nitric acid and the potassium hydroxide solution were both at room temperature. (a) Figure 1 shows part of the thermometer after some potassium hydroxide solution had been added to the dilute nitric acid. What is the temperature shown on the thermometer? The temperature shown is .................... °C Page 1
Transcript
The John Fisher School
Summer revision calculations for AS chemists September 2016
Q1.Dilute nitric acid reacts with potassium hydroxide solution.
The equation for the reaction is:
HNO3 + KOH KNO3 + H2O
A student investigated the temperature change in this reaction.
This is the method the student used.
Step 1 Put 25 cm3 of dilute nitric acid in a polystyrene cup.Step 2 Use a thermometer to measure the temperature of the dilute nitric acid.Step 3 Use a burette to add 4 cm3 of potassium hydroxide solution to the dilute nitric
acid and stir the mixture.Step 4 Use a thermometer to measure the highest temperature of the mixture.Step 5 Repeat steps 3 and 4 until 40 cm3 of potassium hydroxide solution have been
added.
The dilute nitric acid and the potassium hydroxide solution were both at room temperature.
(a) Figure 1 shows part of the thermometer after some potassium hydroxide solution had been added to the dilute nitric acid.
What is the temperature shown on the thermometer?
The temperature shown is .................... °C(1)
(b) Errors are possible in this experiment.
(i) Suggest two causes of random error in the experiment.
(iii) It is difficult to use the data in Figure 2 to find the exact volume of potassium hydroxide solution that would give the maximum temperature.
Suggest further experimental work that the student should do to make it easier to find the exact volume of potassium hydroxide solution that would give the maximum temperature
Concentration = ............................ moles per dm3
(3)
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(e) The student repeated the original experiment using 25 cm3 of dilute nitric acid in a polystyrene cup and potassium hydroxide solution that was twice the original concentration.
She found that:
• a smaller volume of potassium hydroxide solution was required to reach the maximum temperature
• the maximum temperature recorded was higher.
Explain why the maximum temperature recorded was higher.
(iii) The students were asked to comment on the teacher’s observations.
One student said that the observations were not valid because not all the halogens are vapours – chlorine is a gas, bromine is a liquid and iodine is a solid.
(b) A group of students investigated the amount of energy given out when different alcohols are burned. The students used the apparatus shown in the figure below.
In one experiment the temperature of 50 g of water increased from 22.5 °C to 38.3 °C. The mass of alcohol burned was 0.85 g.
(i) The energy used to heat the water, Q, can be found using the equation:
Q = m × c × ΔT
Calculate the heat energy, in joules, given out by burning 0.85 g of the alcohol.
Assume that all of the heat energy given out by burning the alcohol is used to heat the water (specific heat capacity of water = 4.2 J / g / °C).
Heat energy = .................................................. J(2)
(ii) Use your answer to part (b)(i)and the fact that 0.85 g of alcohol were burned to calculate the heat energy that would be given out by 1 g of alcohol.
If you could not answer part (b)(i), use 3000 J as the heat energy given out by burning 0.85 g of the alcohol. This is not the correct answer to part (b)(i).
Heat energy = .................................................. J(1)
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(c) Another group of students investigated the amount of heat energy given out when different alcohols are burned. They used a different, better, set of apparatus than the first group of students.
They used the results from their investigation to calculate the heat energy given out by burning 1 g of each alcohol.
They used a data book to find the theoretical amount of heat energy 1 g of each alcohol should have given out when burned completely.
The students recorded their experimental results and the theoretical values in the table below.
Name of alcohol
Number ofcarbon atoms
inone molecule
ofalcohol
Experimentalamount of heatenergy given
out when 1 g isburned in kJ
Theoreticalamount of
heat energygiven out when1 g is burned
completely in kJ
Methanol 1 11.4 22.7
Ethanol 2 14.5 29.7
Propanol 3 16.0 33.7
Butanol 4 16.8 36.2
Pentanol 5 17.2 37.8
Hexanol 6 17.4 39.1
(i) What is the relationship between the number of carbon atoms in one molecule of the alcohol and the heat energy given out when the alcohol is burned?
Percentage of iron in iron chloride = .................................................%(2)
(Total 6 marks)
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M1.(a) 311
(b) (i) any two from:• incorrect reading of thermometer / temperature• incorrect measurement of volume of acid• incorrect measurement of volume of alkali (burette).
2
(ii) glass is a (heat) conductor or polystyrene is a (heat) insulatoranswer needs to convey idea that heat lost using glass or not lost using polystyreneaccept answers based on greater thermal capacity of glass (such as “glass absorbs more heat than polystyrene”)
1
(c) (i) temperature increases1
(ii) no reaction takes place or all acid used up or potassium hydroxide in excess
1
cool / colder potassium hydroxide absorbs energy or lowers temperatureignore idea of heat energy being lost to surroundings
1
(iii) take more readingsignore just “repeat”
1
around the turning point or between 20 cm3 and 32 cm3
accept smaller ranges as long as no lower than 20 cm3 and no higher than 32 cm3
1
(d) 1.61 or 1.6(12903)correct answer with or without working scores 3if answer incorrect, allow a maximum of two from:moles nitric acid = (2 × 25 / 1000) = 0.05 for 1 markmoles KOH = (moles nitric acid) = 0.05 for 1 markconcentration KOH = 0.05 / 0.031answer must be correctly rounded (1.62 is incorrect)
3
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(e) same amount of energy given out1
which is used to heat a smaller total volume or mixture has lower thermal capacityornumber of moles reacting is the samebut the total volume / thermal capacity is less
if no other marks awarded award 1 mark for idea of reacting faster
1[14]
M2.(a) circle round any one (or more) of the covalent bondsany correct indication of the bond − the line between letters
1
(b) Methane contains atoms of two elements, combined chemically1
(c) (i) activation energy labelled from level of reagents to highest point of curveignore arrowheads
1
enthalpy change labelled from reagents to products
arrowhead must go from reagents to products only1
(ii) 2 O2
1
2 H2Oif not fully correct, award 1 mark for all formulae correct.
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ignore state symbols1
(iii) carbon monoxide is made1
this combines with the blood / haemoglobin or prevents oxygen being carried in the blood / round body or kills you or is toxic or poisonous
dependent on first marking point1
(iv) energy is taken in / required to break bondsaccept bond breaking is endothermic
1
energy is given out when bonds are madeaccept bond making is exothermic
1
the energy given out is greater than the energy taken inthis mark only awarded if both of previous marks awarded
1
(d) (i) energy to break bonds = 1895calculation with no explanation max = 2
1
energy from making bonds = 19981
1895 − 1998 (= −103)orenergy to break bonds = 656energy from making bonds = 759656 − 759 (= −103)
allow:bonds broken − bonds made =413 + 243 − 327 − 432 = -103 for 3 marks.
1
(ii) The C — Br bond is weaker than the C — Cl bond1
[15]
M3.(a) any two from:
• temperature (of the HCl)• mass or length of the magnesium• surface area of the magnesium• volume of HCl
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(b) (i) (a greater concentration has) more particles per unit volumeallow particles are closer together
1
therefore more collisions per unit time or more frequent collisions.1
(ii) particles move fasterallow particles have more (kinetic) energy
1
therefore more collisions per unit time or more frequent collisions1
collisions more energetic (therefore more collisions have energy greater than the activation energy) or more productive collisions
1
(c) (i) add (a few drops) of indicator to the acid in the conical flaskallow any named indicator
1
add NaOH (from the burette) until the indicator changes colour or add the NaOH dropwise
candidate does not have to state a colour change but penalise an incorrect colour change.
1
repeat the titration1
calculate the average volume of NaOH or repeat until concordant
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results are obtained1
(ii) moles of NaOH
0.10 × 0.0272 = 0.00272 molescorrect answer with or without working gains 3 marks
1
Concentration of HCl
0.00272 / 0.005 = 0.544allow ecf from mp1 to mp2
1
correct number of significant figures1
[14]
M4.(a) (i) 10.8 / 24 31.8 / 35.5 57.4 / 16division by atomic mass
1
0.45 0.90 3.59proportion
1
1 : 2 : 8ratio
1
MgCl2O8
formulaif the candidate has just written down the correct answer with no working allow 3 marks. If working is shown award 4 marks.
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allow 2 marks if the candidate has the initial fractions upside down and gets the answer Mg8Cl4O as long as working has been shownallow symbols in any order but do not allow incorrect capitals or subscript and superscript numbers, eg mGCl2O8 (final mark)allow ecf after step 2 provided a sensible attempt has been made in step 1
1
(ii) because it has electrostatic forces of attraction or attraction between oppositely charged ions
any mention of covalent bonds or molecules or intermolecular forces max. 2 marks
1
giant structure / lattice or many bondsaccept attraction between positive and negative ionsdo not accept atoms
1
that need lots of energy to overcome / break1
(b) any three from:
• more can be stored in a given volume or takes up less spacecould be answered in terms of comparing liquid with gas, but not simply the disadvantages of storing a gaseous fuel
• (liquid has) greater energy density• (liquid) does not require high pressure containers• (liquid is) less likely to leak
allow easier to detect a leak3
[10]
M5. (a) (i) chlorine > bromine > iodineaccept symbols in place of names (Cl2 or Cl, etc)
1
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(ii) any two suitable comparisons about the extent to which the iron wool glowed
eg chlorine is more reactive than bromine because iron glowed morebrightly with chlorine than bromine
eg bromine is more reactive than iodine because iron glowed withbromine but not with iodine
eg iodine is the least reactive because it is the only one that did not glow
eg chlorine is the most reactive as it was the only one that glowed brightlyor glowed most brightly
2 max.
(iii) NO (no marks)accept YES (no marks)
because the observations were made using the halogen vapoursbecause the observations were made using the halogen vapours
1
so the fact that the three elements are naturally solid, liquid or gas isnot relevant
so the iodine (and bromine) must have been heated1
(b) division of masses by Ar valuesif division inverted (Ar / mass) then award zero marks
1
correct answers of divisionif one slip in Ar values used, do not award first mark, but award remainder consequentially
1
simplification by division by smallest1
correct formula (NaClO3)
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accept elements in any order (such as ClNaO3)correct answer alone with or without working = 4 marks
1[9]
M6.Divide by Ar:
Na = 22.8 / 23
B = 21.8 / 11
O = 55.4 / 16if student has calculated moles upside down they can score mp 3 mp 4 and mp 5 as follows:Na 23 / 22.8B 11 / 21.8O 16 / 55.4
1
Values
0.9911.01
1.980.505
3.460.289
1
Divide by the smallest
1 : 2 : 3.5Divide by the smallest (1)3.5 : 1.75 : 1
1
Whole number ratio
2 : 4 : 7
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Whole number ratio (1)14 : 7 : 4
1
Empirical formula
Na2B4O7
Empirical formula (1)Na14B7O4
if no working shown allow 4 marks for Na2B4O7
1[5]
M7.(a) (i) redignore pink
1
(ii) add silver nitrate (solution)1
white precipitatedependent on addition of silver nitrateignore addition of another acidif hydrochloric acid added max 1 mark
1
(b) suitable named alkali / sodium hydroxide solution in burette1
add alkali solution until (indicator) becomes pink / red1
if acid to acid titration described, first two marking points not available
any two from:
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• wash / rinse equipment• add dropwise or slowly (near end point)• swirl / mix• read (meniscus) at eye level• white background• read start and final burette levels / calculate the volume needed• repeat
2
(c) does not ionise / dissociate completelyallow for acids of the same concentration, weak acids have a higher pH or fewer hydrogen ions
1
(d) (i) ring round COOH
1
(ii)
if not fully correct, allow 1 mark for correct ester group − minimum
2
[11]
M8.(a) (i)
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allow –OH
1
(ii) (C2H5OH +) 3O2 ➔ 2CO2 + 3H2O
formulaeall 3 formulae gain 2 marks;2 formulae gain 1 mark
2
balanceddependent on formulaeaccept multiples
1
(iii) energy is required to break bonds1
energy is released when bonds are formed1
exothermic because more energy is released than taken in1
(b) (i) 3318 or 3320 (J)correct answer with or without working gains 2 marksif answer incorrect, allow 1 mark for correct working, eg:temperature change = (38.3 – 22.5 = ) 15.8 or Q = (50 x 4.2 x 15.8 =)allow ecf from temp change
2
(ii) if answer from part (i) used:
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(3318 / 0.85) = 3904allow ecf from part (i)
if value of 3000 used:
3000 / 0.85 = 3529accept 2 sig figs up to calculator value
1
(c) (i) the more carbon atoms in one molecule, the more heat energy is given out
do not accept proportional1
(ii) heat loss (to surroundings or to the can or to the air)accept not all energy used to heat waterallow incomplete combustionallow not all burnedignore thermometer or balance errors
1
(iii) incomplete combustion (occurs when more C atoms in molecule)1
(iv) 1370correct answer with or without working gains 3 marks(Mr ethanol =) 46 gains 1 mark46 x 29.7 gains 2 marksmax 2 marks if incorrect sig figs (1366.2, 1366, 1400)final answer ecf on incorrect Mr max 2 marks
3[16]
M9.(a) (i) an alloy1
(ii) harder1
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(b) (i) 162.5correct answer with or without working gains 2 marksif no answer or incorrect answer then evidence of correct working [56 + (3x35.5)] gains 1 mark
2
(ii) 34.46accept rounding from 34 - 34.5correct answer with or without working gains 2 marksaccept ecf from (b)(i) correctly calculated for 2 marksif no answer or incorrect answer then evidence of 56 / 162.5 or 56 / answer to (b)(i) gains 1 mark
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