revision exercise (a) answers section b

Revision Exercise (a)

Moles, Redox and GasesSection B

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

Calculations are required.

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

Calculations are required.

We cannot assume that the increase in volume will cause it to burst.

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

Calculations are required.

We cannot assume that the increase in volume will cause it to burst.

Write equations and substitute correctly with the correct units.

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

p Vf = n R Tf

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

At 5 C

975 cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

At 5 C

975 cm3

At 25 C

? cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

p Vf = n R Tf

Vf / Tf = (n R) / p

At 25 C

? cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

p Vf = n R Tf

Vf / Tf = (n R) / p

We have 2 unknowns, n & p

At 25 C

? cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

p Vf = n R Tf

Vf / Tf = (n R) / p

We have 2 unknowns, n & p

We can find them out. This is how.

At 25 C

? cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

At 5 C

975 cm3

p Vi = n R Ti

Vi / Ti = (n R) / p

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

At 5 C

975 cm3

p Vi = n R Ti

Vi / Ti = (n R) / p

p Vf = n R Tf

Vf / Tf = (n R) / p

At 25 C

? cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

At 5 C

975 cm3

Vi / Ti = Vf / Tf

Vf = (Vi Tf) / Ti

Substitute the correct values in the correct units.

At 25 C

? cm3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

Vi / Ti = Vf / Tf

Vf = (Vi Tf) / Ti

Substitute the correct values in the correct units.

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

Vi / Ti = Vf / Tf

Vf = (Vi Tf) / Ti

Substitute the correct values in the correct units.

Vf = 975 X 10-6 ( 25 + 273) / (5 + 273)

= 1045 X 10-6 m3

1. A balloon can hold 1000 cm3 of air before bursting. The balloon contains 975 cm3 of air at 5 C. Will it burst when it is taken into a house at 25 C? Assume that the pressure of the gas in the balloon remains constant.

[2]

Vi / Ti = Vf / Tf

Vf = (Vi Tf) / Ti

Substitute the correct values in the correct units.

Vf = 975 X 10-6 ( 25 + 273) / (5 + 273)

= 1045 X 10-6 m3

It will BURST!

1 a) What do you understand by the term ideal gas. [1]b) State two assumptions made of the kinetic theory of gases. [2]c) List two factors that cause real gases to deviate from ideal gas

behaviour. [1] d) Under what conditions of temperature and pressure do real gases

behave most ideally? Give reasons for your answers. [2]e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder

is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]Assume that the hydrocarbons in gasoline have an average molecular mass of 100 and react with oxygen in a 1:12 mole ratio and the pressure change is negligible after injection.

ii) What is the mass of gasoline that needs to be injected into the cylinder for complete reaction with the hydrocarbons.

[2]

1 a) What do you understand by the term ideal gas. [1]

1 a) What do you understand by the term ideal gas. [1]

Ideal gas is a gas that obeys the assumptions of the kinetic theory of gases.

2 b) State two assumptions made of the kinetic theory of gases. [2]

2 b) State two assumptions made of the kinetic theory of gases. [2]

Any two

The particles of gas are in constant, random motion and collide with each other and the walls of the container.

The volume of the gas particles are negligible as compared to the volume of the gas.

Collisions of the particles with each other and the walls of the container are perfectly elastic.

There are no intermolecular forces between molecules of gas.

2 c) List two factors that cause real gases to deviate from ideal gas behaviour. [1]

2 c) List two factors that cause real gases to deviate from ideal gas behaviour. [1]

Any 2

Temperature

Pressure

Nature of Gas

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

High Temperature and Low Pressure

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

High Temperature

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

High Temperature

At high temperatures, gas particles possess more kinetic energy and thus are able to overcome the intermolecular forces between them. This will allow them to behave more ideally.

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

High Temperature

At high temperatures, gas particles possess more kinetic energy and thus are able to overcome the intermolecular forces between them. This will allow them to behave more ideally.

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

Low Pressure

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

Low Pressure

At low pressures, gas molecules are not as closely packed, therefore the volume of the gas particles are negligible as compared to the volume of the gas.

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

Low Pressure

At low pressures, gas molecules are not as closely packed, therefore the volume of the gas particles are negligible as compared to the volume of the gas.

At low pressures, gas molecules are further apart and therefore do not form intermolecular bonds as easily. Thus behaving more

ideally.

2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

Low Pressure

At low pressures, gas molecules are not as closely packed, therefore the volume of the gas particles are negligible as compared to the volume of the gas.

At low pressures, gas molecules are further apart and therefore do not form intermolecular bonds as easily. Thus behaving more ideally.

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

We can find the no. of moles of air so 21% of it is oxygen.

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

101 kPa = 101 000 Pa

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

101 kPa = 101 000 Pa

500 cm3 = 0.0005 m3

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

101 kPa = 101 000 Pa

500 cm3 = 0.0005 m3

60 0C = 333 K

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

101 kPa = 101 000 Pa

500 cm3 = 0.0005 m3

60 0C = 333 K

p V = n R T

where n is the number of moles of air

n = ( p V / R T)

= ( 101 000 X 0.0005 ) / 8.314 X 333

= 0.01824 mol

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

p V = n R T

where n is the number of moles of air

n = ( p V / R T)

= ( 101 000 X 0.0005 ) / 8.314 X 333

= 0.01824 mol

21% of n(Air) = n(O2)

2 e) A Volvo engine has a cylinder volume of about 500 cm3. The cylinder is full of air at 60 oC and a pressure of 101 kPa. i) Calculate the number of moles of oxygen in the cylinder.

(% composition by volume of oxygen in air = 21) [2]

p V = n R T

where n is the number of moles of air

n = ( p V / R T)

= ( 101 000 X 0.0005 ) / 8.314 X 333

= 0.01824 mol

21% of n(Air) = n(O2)

n(O2) = 0.01824 X 21%

= 0.00383 mol

2 Assume that the hydrocarbons in gasoline have an average molecular mass of 100 and react with oxygen in a 1:12 mole ratio and the pressure change is negligible after injection.

ii) What is the mass of gasoline that needs to be injected into the cylinder for complete reaction with the hydrocarbons.

[2]

2 Assume that the hydrocarbons in gasoline have an average molecular mass of 100 and react with oxygen in a 1:12 mole ratio and the pressure change is negligible after injection.

ii) What is the mass of gasoline that needs to be injected into the cylinder for complete reaction with the hydrocarbons.

[2]

n(O2) : n(Hydrocarbon)

12 : 1

2 Assume that the hydrocarbons in gasoline have an average molecular mass of 100 and react with oxygen in a 1:12 mole ratio and the pressure change is negligible after injection.

ii) What is the mass of gasoline that needs to be injected into the cylinder for complete reaction with the hydrocarbons.

[2]

n(O2) : n(Hydrocarbon)

12 : 1

It is a mixture of hydrocarbons so u cannot write like CxHy

2 Assume that the hydrocarbons in gasoline have an average molecular mass of 100 and react with oxygen in a 1:12 mole ratio and the pressure change is negligible after injection.

ii) What is the mass of gasoline that needs to be injected into the cylinder for complete reaction with the hydrocarbons.

[2]

n(O2) : n(Hydrocarbon)

12 : 1

It is a mixture of hydrocarbons so u cannot write like CxHy

n(hydrocarbons) = 0.00383 mol / 12

= 3.192 X 10-4 mol

2 Assume that the hydrocarbons in gasoline have an average molecular mass of 100 and react with oxygen in a 1:12 mole ratio and the pressure change is negligible after injection.

ii) What is the mass of gasoline that needs to be injected into the cylinder for complete reaction with the hydrocarbons.

[2]

n(O2) : n(Hydrocarbon)

12 : 1

It is a mixture of hydrocarbons so u cannot write like CxHy

n(hydrocarbons) = 0.00383 mol / 12

= 3.192 X 10-4 mol

mass = Mr X mol

= 100 X 3.192 X 10-4

= 0.0319 g

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

a) Write an ionic equation for the neutralization reaction. [1]b) Calculate the number of moles of X2O6 dissolved in water.

[2]c) Calculate the relative atomic mass of X and predict the

identity of X. [3]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

a) Write an ionic equation for the neutralization reaction. [1]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

a) Write an ionic equation for the neutralization reaction. [1]

Acid Base Reaction.

H+ (aq) + OH- (aq) H2O (aq)

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

H+ + OH- H2O

n(OH-) = (52.3/1000) X 0.100= 0.00523 mol

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

H+ + OH- H2O

n(OH-) = (52.3/1000) X 0.100= 0.00523 mol

n(H+) = n(OH-)= 0.00523 mol

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

H+ + OH- H2O

n(OH-) = (52.3/1000) X 0.100= 0.00523 mol

n(H+) = n(OH-)= 0.00523 mol

3 H+ : 1 X2O6

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

b) Calculate the number of moles of X2O6 dissolved in water.[2]

H+ + OH- H2O

n(OH-) = (52.3/1000) X 0.100= 0.00523 mol

n(H+) = n(OH-)= 0.00523 mol

3 H+ : 1 X2O6

n(X2O6) = 0.00523 / 3= 1.74 X 10-3 mol

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

c) Calculate the relative atomic mass of X and predict the identity of X. [3]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

c) Calculate the relative atomic mass of X and predict the identity of X. [3]

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

c) Calculate the relative atomic mass of X and predict the identity of X. [3]

Mr of X2O6 = Mass / Mole= 0.47 / (1.74 X 10-3)= 270

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

c) Calculate the relative atomic mass of X and predict the identity of X. [3]

Mr of X2O6 = Mass / Mole= 0.47 / (1.74 X 10-3)= 270

Mr of X = [270 – (6 X 16)] / 2= 87

3. When 1 mole of X2O6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X2O6 is dissolved in water and the resulting solution required 52.3 cm3 of 0.100 mol dm-3 sodium hydroxide solution for complete neutralization.

c) Calculate the relative atomic mass of X and predict the identity of X. [3]

Mr of X2O6 = Mass / Mole= 0.47 / (1.74 X 10-3)= 270

Mr of X = [270 – (6 X 16)] / 2= 87

X is Strontium