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A LEVEL CHEMISTRY OCR A PAPER 3

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Q12 b] The structure of compound A is shown below. [refer to the PMT June 2017 A LEVEL CHEMISTRY OCR A Paper 3 QP for the structure of compound A] Using this information and the student's results, determine the molar mass of A and the formula of the alkyl group R. Show all your working. answer1. The mean titre is 22.50 cm3 of 0.0840 mol dm-3 NaOH moles = conc. x vol. so 0.084 x 0.0225 = 1.89 x10^-3 moles of NaOH used 2. molar ratio of NaOH : A is 1 : 1 [as A is a monobasic acid], so there are 1.89 x10^-3 moles of A used as well 3. conc. = moles / vol, so 1.89 x10^-3 / 0.025 = a conc. of 0.084 mol dm-3 for A

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A LEVEL
CHEMISTRY
OCR A PAPER 3

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, Q1 >>> Within the permafrost in Arctic regions of the Earth, large amounts of methane are trapped
within ice as 'methane hydrate', CH4•xH2O. Methane makes up about 13.4% of the mass of 'methane
hydrate'.

Scientists are concerned that global warming will melt the permafrost, releasing large quantities of
methane into the atmosphere.

The H-O-H bond angle in ice is about 109° but about 105° in gaseous H2O. Explain why there is this
difference [3 marks] answerThere are hydrogen bonds between water molecules in ice: 2 bonded pairs
and 2 lone pairs of electrons

However, there is no hydrogen bonds between water molecules of gaseous H2O so the lone pairs of
electrons repel the bonded pairs even more



Q2 >>> Why are scientists concerned about the release of methane into the atmosphere ?
answerMethane is a greenhouse gas that causes global warming



Q3 >>> Determine the formula of 'methane hydrate', CH4•xH2O. In the formula, show the value of x to
two decimal places. answer1. The Mr of CH4 is 16.0 and the Mr of H2O is 18.0

2. In 100g of CH4•xH2O there is 13.4g of CH4, and so there must be 86.6g of H2O

3a]. 13.4g of CH4 is 0.8375 moles of CH4

3b]. 86.6g of H2O is 4.81111111111 moles of H2O

4] divide everything by the smallest number of moles

5.74461028192 = moles of H2O

The formula of methane hydrate is: *CH4 •5.74H2O*



Q4 >>> Calculate the volume of methane, in dm3, that would be released from the melting of each 1.00
kg of 'methane hydrate' at 101 kPa and 0 °C. Give your answer to three significant figures. answer1. PV =
nRT so V = nRT/P

2. P = Pa, V = M3, T = K, R = gas constant, n = moles

3. 101kPa, 101,000Pa, 0 °C = 273K

4. Mr of methane hydrate = 119.32 g mol-1

5. 1.00kg of methane hydrate = 1000g of methane hydrate, so

mass / Mr of methane hydrate = 8.38082467315 moles of methane hydrate

6. nRT / P = [8.38082467315 x 8.314 x 273] / 101,000 = 0.18833804097m3

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