The document provides step-by-step working of a chemistry problem calculating the volume of gas in a balloon after a temperature change. It starts with the information that a balloon contains 975 cm3 of air at 5°C and asks if it will burst when brought inside at 25°C, assuming constant pressure. The solution uses the ideal gas law to set up two equations relating the initial and final states, and solves them to find the final volume is 1045 cm3, so the balloon will burst as it exceeds the maximum volume of 1000 cm3.

1. Revision Exercise (a) Moles, Redox and Gases Section B

2. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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]

3. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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.

4. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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.

5. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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.

6. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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]

7. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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]

8. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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]

9. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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]

10. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 V f = n R T f

11. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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]

12. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 cm 3

13. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 cm 3 At 25  C ? cm 3

14. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 V f = n R T f V f / T f = (n R) / p At 25  C ? cm 3

15. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 V f = n R T f V f / T f = (n R) / p We have 2 unknowns, n & p At 25  C ? cm 3

16. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 V f = n R T f V f / T f = (n R) / p We have 2 unknowns, n & p We can find them out. This is how. At 25  C ? cm 3

17. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 V i = n R T i V i / T i = (n R) / p At 5  C 975 cm 3

18. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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 V i = n R T i V i / T i = (n R) / p p V f = n R T f V f / T f = (n R) / p At 5  C 975 cm 3 At 25  C ? cm 3

19. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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] V i / T i = V f / T f V f = (V i T f ) / T i Substitute the correct values in the correct units. At 5  C 975 cm 3 At 25  C ? cm 3

20. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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] V i / T i = V f / T f V f = (V i T f ) / T i Substitute the correct values in the correct units.

21. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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] V i / T i = V f / T f V f = (V i T f ) / T i Substitute the correct values in the correct units. V f = 975 X 10 -6 ( 25 + 273) / (5 + 273) = 1045 X 10 -6 m 3

22. 1. A balloon can hold 1000 cm 3 of air before bursting. The balloon contains 975 cm 3 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] V i / T i = V f / T f V f = (V i T f ) / T i Substitute the correct values in the correct units. V f = 975 X 10 -6 ( 25 + 273) / (5 + 273) = 1045 X 10 -6 m 3 It will BURST!

26. 2 b) State two assumptions made of the kinetic theory of gases. [2]

27. 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.

28. 2 c) List two factors that cause real gases to deviate from ideal gas behaviour. [1]

29. 2 c) List two factors that cause real gases to deviate from ideal gas behaviour. [1] Any 2 Temperature Pressure Nature of Gas

30. 2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2]

31. 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

32. 2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2] High Temperature

33. 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.

34. 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 .

35. 2 d) Under what conditions of temperature and pressure do real gases behave most ideally? Give reasons for your answers. [2] Low Pressure

36. 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.

37. 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.

38. 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.

39. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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]

40. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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.

41. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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

42. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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 cm 3 = 0.0005 m 3

43. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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 cm 3 = 0.0005 m 3 60 0 C = 333 K

44. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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 cm 3 = 0.0005 m 3 60 0 C = 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

45. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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(O 2 )

46. 2 e) A Volvo engine has a cylinder volume of about 500 cm 3 . The cylinder is full of air at 60 o C 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(O 2 ) n(O 2 ) = 0.01824 X 21% = 0.00383 mol

47. 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]

48. 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(O 2 ) : n(Hydrocarbon) 12 : 1

49. 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(O 2 ) : n(Hydrocarbon) 12 : 1 It is a mixture of hydrocarbons so u cannot write like C x H y

50. 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(O 2 ) : n(Hydrocarbon) 12 : 1 It is a mixture of hydrocarbons so u cannot write like C x H y n(hydrocarbons) = 0.00383 mol / 12 = 3.192 X 10 -4 mol

51. 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(O 2 ) : n(Hydrocarbon) 12 : 1 It is a mixture of hydrocarbons so u cannot write like C x H y n(hydrocarbons) = 0.00383 mol / 12 = 3.192 X 10 -4 mol mass = M r X mol = 100 X 3.192 X 10 -4 = 0.0319 g

52. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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 X 2 O 6 dissolved in water. [2] c) Calculate the relative atomic mass of X and predict the identity of X. [3]

53. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. a) Write an ionic equation for the neutralization reaction. [1]

54. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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)  H 2 O (aq)

55. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2]

56. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2]

57. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2]

58. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2] H + + OH -  H 2 O n(OH - ) = (52.3/1000) X 0.100 = 0.00523 mol

59. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2] H + + OH -  H 2 O n(OH - ) = (52.3/1000) X 0.100 = 0.00523 mol n(H + ) = n(OH - ) = 0.00523 mol

60. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2] H + + OH -  H 2 O n(OH - ) = (52.3/1000) X 0.100 = 0.00523 mol n(H + ) = n(OH - ) = 0.00523 mol 3 H + : 1 X 2 O 6

61. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 of 0.100 mol dm -3 sodium hydroxide solution for complete neutralization. b) Calculate the number of moles of X 2 O 6 dissolved in water. [2] H + + OH -  H 2 O n(OH - ) = (52.3/1000) X 0.100 = 0.00523 mol n(H + ) = n(OH - ) = 0.00523 mol 3 H + : 1 X 2 O 6 n(X 2 O 6 ) = 0.00523 / 3 = 1.74 X 10 -3 mol

62. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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]

63. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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]

64. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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] M r of X 2 O 6 = Mass / Mole = 0.47 / (1.74 X 10 -3 ) = 270

65. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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] M r of X 2 O 6 = Mass / Mole = 0.47 / (1.74 X 10 -3 ) = 270 M r of X = [270 – (6 X 16)] / 2 = 87

66. 3. When 1 mole of X 2 O 6 is dissolved in water, 3 moles of hydrogen ions is liberated. 0.47 g of X 2 O 6 is dissolved in water and the resulting solution required 52.3 cm 3 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] M r of X 2 O 6 = Mass / Mole = 0.47 / (1.74 X 10 -3 ) = 270 M r of X = [270 – (6 X 16)] / 2 = 87 X is Strontium