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ce07 mc chem
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33. The answer is B. According to Avogadro’s law, under constant temperature and pressure, the volume ratio of gases is equal to their mole ratio. 2CO(g) + O2(g) → 2CO2(g) Volume ratio (CO used) : (O2 used) = 2 : 1 When 50 cm3 of CO is completely reacted, volume of O2 used = 50 x (1/2) = 25 cm3 Hence, CO is the limiting reactant (completely reacted). Volume of O2 unreacted = 50 - 25 = 25 cm3 Volume ratio (CO used) : (CO2 formed) = 2 : 2 = 1 : 1 Volume of CO reacted = 50 cm3 Volume of CO2 formed = 50 cm3 Final volume at the end of the combustion = (Volume of O2 unreacted) + (Volume of CO2 formed) = 25 + 50 = 75 cm3 ========== 38. The answer is D. A is incorrect. Pb(s) + H2SO4(aq) → PbSO4(s) + H2(g) Since Pb is unreactive, the reaction is very slow. Furthermore, the PbSO4(s) formed will coat on the surface of Pb(s), and this will stop the reaction soon after the reaction starts. B is incorrect. Pb(s) + CuSO4(aq) → PbSO4(s) + Cu(s) The PbSO4(s) are mixed with Cu(s), and it is difficult to separate them. Further, PbSO4(s) and Cu(s) formed will coat on the surface of Pb(s), and this may stop the reaction soon after the reaction starts. C is incorrect. PbO(s) + H2SO4(aq) → PbSO4(s) + H2O(l) The PbSO4(s) formed will coat on the surface of PbO(s), and this will stop the reaction soon after the reaction starts. D is correct. Pb(NO3)2(aq) + H2SO4(s) → PbSO4(s) + 2HNO3(aq) or Pb2+(aq) + SO42-(aq) → PbSO4(s) When the solutions are mixed, PbSO4(s) is precipitated out. It is easy to obtain PbSO4(s) by filtration of the reaction mixture, followed by washing and drying the residue.
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ce07 mc chem
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33. 50 cm3 of carbon monoxide burns completely in 50 cm3 of oxygen . All the volumes are measured at room temperature and pressure. The final gaseous volume at the end of the combustion isA. 50 cm3B. 75 cm3 C. 100 cm3D. 150 cm338.Which methods given below is most suitable to prepare a sample of... 顯示更多 33. 50 cm3 of carbon monoxide burns completely in 50 cm3 of oxygen . All the volumes are measured at room temperature and pressure. The final gaseous volume at the end of the combustion is A. 50 cm3 B. 75 cm3 C. 100 cm3 D. 150 cm3 38. Which methods given below is most suitable to prepare a sample of lead(II) sulphate? A. Add Lead to dilute sulphuric acid. B. Add Lead to copper(II) sulphate solution. C. Add lead(II) oxide to dilute sulphuric acid. D. Add lead(II) nitrate solution to dilute sulphuric acid. 更新: please explain with detail thanks~~最佳解答:
33. The answer is B. According to Avogadro’s law, under constant temperature and pressure, the volume ratio of gases is equal to their mole ratio. 2CO(g) + O2(g) → 2CO2(g) Volume ratio (CO used) : (O2 used) = 2 : 1 When 50 cm3 of CO is completely reacted, volume of O2 used = 50 x (1/2) = 25 cm3 Hence, CO is the limiting reactant (completely reacted). Volume of O2 unreacted = 50 - 25 = 25 cm3 Volume ratio (CO used) : (CO2 formed) = 2 : 2 = 1 : 1 Volume of CO reacted = 50 cm3 Volume of CO2 formed = 50 cm3 Final volume at the end of the combustion = (Volume of O2 unreacted) + (Volume of CO2 formed) = 25 + 50 = 75 cm3 ========== 38. The answer is D. A is incorrect. Pb(s) + H2SO4(aq) → PbSO4(s) + H2(g) Since Pb is unreactive, the reaction is very slow. Furthermore, the PbSO4(s) formed will coat on the surface of Pb(s), and this will stop the reaction soon after the reaction starts. B is incorrect. Pb(s) + CuSO4(aq) → PbSO4(s) + Cu(s) The PbSO4(s) are mixed with Cu(s), and it is difficult to separate them. Further, PbSO4(s) and Cu(s) formed will coat on the surface of Pb(s), and this may stop the reaction soon after the reaction starts. C is incorrect. PbO(s) + H2SO4(aq) → PbSO4(s) + H2O(l) The PbSO4(s) formed will coat on the surface of PbO(s), and this will stop the reaction soon after the reaction starts. D is correct. Pb(NO3)2(aq) + H2SO4(s) → PbSO4(s) + 2HNO3(aq) or Pb2+(aq) + SO42-(aq) → PbSO4(s) When the solutions are mixed, PbSO4(s) is precipitated out. It is easy to obtain PbSO4(s) by filtration of the reaction mixture, followed by washing and drying the residue.
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