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When 3 cm 3 of ethylene and 12 cm 3 of hydrogen are mixed

When 3 cm 3 of ethylene and 12 cm 3 of hydrogen are mixed under appropriate conditions, they react to form ethane: C 2 H 4(g) + H 2(g) C 2 H 6(g) What is the final volume of the reaction mixture ?. When 3 cm 3 of ethylene and 12 cm 3 of hydrogen are mixed

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When 3 cm 3 of ethylene and 12 cm 3 of hydrogen are mixed

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  1. When 3 cm3 of ethylene and 12 cm3 of hydrogen are mixed under appropriate conditions, they react to form ethane: C2H4(g) + H2(g) C2H6(g) What is the final volume of the reaction mixture?

  2. When 3 cm3 of ethylene and 12 cm3 of hydrogen are mixed under appropriate conditions, they react to form ethane: C2H4(g) + H2(g) C2H6(g) What is the final volume of the reaction mixture? Gay Lussac 3 cm3 of ethylene reacts with 3 cm3 of hydrogen to give 3 cm3 of ethane 9 cm3 of hydrogen remain so the final volume = 12 cm3

  3. In the reaction Cu + 2 HNO3 Cu(NO3)2 + H2(g) what volume of 10 M HNO3 is required to produce 11.2 dm3 of hydrogen?

  4. In the reaction Cu + 2 HNO3 Cu(NO3)2 + H2(g) what volume of 10 M HNO3 is required to produce 11.2 dm3 of hydrogen? 11.2 dm3 11.2/22.4 = 0.5 moles of H2 2 molecules of HNO3 produce 1 molecule of H2 2 moles of HNO3 produce 1 mole of H2 => 1 mole of HNO3 is required 10 M HNO3  10 moles HNO3 per Litre 10 moles in 1 L => need 1/10 or 100 ml of HNO3

  5. What mass of solid sodium hydroxide, of 97% purity by weight, would be required to prepare 0.5 dm3 of a 0.46 M solution? Molar mass of NaOH = 23 + 16 + 1 = 40 g 0.5 dm3 of 1 M solution would require 0.5 moles 0.5 dm3 of 0.46 M solution would require 0.5 x 0.46 moles 0.5 dm3 of 0.46 M solution would require 0.5 x 0.46 x 40 g 0.5 dm3 of 0.46 M solution (97 % purity) would require 0.5 x 0.46 x 40/0.97 g = 9.48 g

  6. Combustion of 1g of menthol, the flavouring agent obtained in peppermint oil, yielded 1.161 g of H2O and 2.818 g of CO2. Given that menthol contains carbon, hydrogen and oxygen, what is its empirical formula? Hydrogen: 1.161 g of H2O  1.161/(2 + 16) = 0.0645 moles of water => 0.0645 x 2 = 0.129 moles of hydrogen Carbon: 2.818 g of CO2 2.818/(12 + 2 x 16) = 0.0640 moles of CO2 => 0.0640 of moles of carbon

  7. Oxygen: 1.0 - 0.13 - 0.768 g = 0.102 g of oxygen  0.102/16 = 0.064 moles of oxygen 0.129 H : 0.064 C : 0.064 O 2 H : 1 C : 1 O Empirical formula: CH2O

  8. The concentration of glucose (C6H12O6) in blood is normally 90 mg per 100 ml. What is the molarity of glucose? Molar mass of glucose is 6 x 12 + 12 x 1.008 + 6 x 16 = 180.1 g mol-1 90 mg = 9 x 10-2 g => molarity = 9 x 10-2 g 180.1 x 0.1 g mol-1 dm3 = 5 mM

  9. What is the oxidation state of arsenic in H2AsO4-? O.S. of H = 1 O.S. of O = -2 sum of oxidation states = -1 = 2 (1) + x + 4(-2) = x - 6 => x = +5

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