F321 Module 1 Practice 4 Answers

7/27/2019 F321 Module 1 Practice 4 Answers

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F321 module 1 Practice 4:

1. (i) Molar mass CaO = 56.1 (g mol1

) (anywhere) 2

moles CaO = 1.56

50.1

= = 0.0267/0.027 calc: 0.0267379

Allow 56 which gives 0.0268

(ii) moles HNO3

= 2 0.0267

= 0.0534 or 0.0535 /0.053 mol

(i.e. answer to (i) x 2)

volume of HNO3

= 2.50

10005)(or0.0534

= 21.4 cm3

2

calc from value above = 21.3903743

If 0.053 mol, answer is 21 cm3

but accept 21.2 cm3

If 0.054 mol, answer is 22 cm3

but accept 21.6 cm3

[4]

2. (i) dative covalent, bonded pair comes from same atom/ 1

electron pair is donated from one atom/

both electrons are from the same atom

(ii) Ca(NO3)2 CaO + 2NO

2+ O

2

or double equation with 2/2/4/1 1[2]

3. (i) 203.3 g mol1

1

Accept 203

(ii) white precipitate / goes white 1

(iii) Ag+(aq) + Cl

(aq) AgCl(s)

equation

state symbols 2

AgCldissolves in NH3(aq)

(iv) AgBr dissolves in conc NH3(aq)/

partially soluble in NH3(aq)

AgI insoluble in NH3

(aq) 3

[7]

The King's CE School 1


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4. (i) a proton donor 1

(ii) MgO + 2HCl MgCl2+ H

2O 1

[2]

5.

isotope protons neutrons electrons

12C 6 6 6

13C 6 7 6

[2]

6. (i) mass spectrometry 1

(ii) mass of an isotope compared with carbon-12

1/12th of mass of carbon-12/on a scale where carbon-12 is

12 2

mass of 1 mole of the isotope/mass of 1 mole of carbon-12

is equivalent to the first mark

mass of the isotope that contains the same number of

atoms as are in 1 mole of carbon-12

1 mark (marklost because of mass units)

(iii) 12 95/100 + 13 5/100 OR 12.05

= 12.1 (mark for significant figures)

(12.1 scores both marks) 2[5]

7. (i) moles CO2

= 1000 /44 mol = 22.7 mol

volume CO2 in 2000 = 22.7 24 = 545 dm3

(ii) reduction = 545 60/100 = 327 dm3

[3]

8. (a) Ca(s) + ..2 HCl(aq) .CaCl2(aq) + .H

2(g). 2

(g) not required for H2

(b) In Ca, oxidation state = 0 and 2

In CaCl2, oxidation state = +2



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9. (i) moles HCl= 2.0 50/1000 = 0.10 1

(ii) moles Ca = moles HCl= 0.050

mass Ca = 40.1 0.050 = 2.00 g / 2.005 g 2

(accept 40 0.050 = 2.0 g)

(mass Ca of 4.0 g would score 1 mark as ecf as molar ratio

has not been identified)

(iii) Ca has reacted with water

Ca + 2H2O Ca(OH)

2+ H

2

state symbols not required

1st mark for H2

3

2nd mark is for the rest of the balanced equation[6]

10. (a) (i) atoms of same element/same atomic number.. with

different numbers of neutrons/different masses 1

(ii) isotope protons neutrons electrons 2

46Ti 22 24 22

47Ti 22 25 22

(b)724.47/

10081.3)(489.8)(478.9)(46 ++

=r

A

= 47.7 2[5]


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F321 Module 1 Practice 4 Answers