標題:
F.4 CHEM
1. 5.0g of copper is burnt in air.a)find the mass of oxide formed.b)the experimental mass of oxide formed is usually less than the theoretical answer in(a).explain why.2. 24.6g of calcium was added into a beaker of water. Determine the mass of gas evolved.3. determine the number of atoms in the... 顯示更多 1. 5.0g of copper is burnt in air. a)find the mass of oxide formed. b)the experimental mass of oxide formed is usually less than the theoretical answer in(a).explain why. 2. 24.6g of calcium was added into a beaker of water. Determine the mass of gas evolved. 3. determine the number of atoms in the following compound. a)64g of argon gas help me please!!thanks..
最佳解答:
1. a) 2Cu + O2 → 2CuO Molar ratio Cu : CuO = 2 : 2 = 1 : 1 Molar mass of Cu = 63.5 g/mol Molar mass of CuO = 63.5 + 16 = 79.5 No. of moles of Cu reacted = 5/63.5 = 0.0787 mol No. of moles of CuO formed = 0.0787 mol Mass of CuO formed = 0.0787 * 79.5 = 6.26 g ...... (Answer) b) The copper contains some impurities which does not reacted. Or: The copper is not completely reacted. 2. Ca + 2H2O → Ca(OH)2 + H2 Mole ratio Ca : H2 = 1 : 1 Molar mass of Ca = 40 g/mol Molar mass of H2 = 1*2 = 2 g/mol No. of moles of Ca used = 24.6/40 = 0.615 mol No. of moles of H2 formed = 0.615 mol Mass of H2 formed = 0.615 * 2 = 1.23 g ...... (Answer) 3. Molar mass of Ar = 40 g/mol No. of moles of Ar = 64/40 = 1.6 mol No. of Ar atoms = 1.6 * (6.02 * 1023) = 9.63 * 1023 ...... (Answer)
其他解答:
1a. 2Cu(s) + O2(g) --> 2CuO(s) mole of Cu = 5/63.5 from the equation, 2 mole Cu produced 2 mole CuO then 5/63.5 mole Cu produced 5/63.5 mole CuO therefore, mass of oxide formed = 5/63.5 x (63.5+16) = 6.26g 1b. There may have loss of material during the experiment 2. Ca + H2O --> CaO + H2 mole of Ca used = 24.6/40.1 from the equation, 1 mole Ca form 1 mole H2 24.6/40.1 mole Ca form 24.6/40.1 mole H2 therefore, mass of H2 evolved = 24.6/40.1 x (1x2) = 1.23g 3. no.of mole of 64g Ar = 64/40 = 1.6 By the Avogadro Constant, 1 mole contains 6.02 x 10^23 atom therefore, no. of atoms that Ar contained = 1.6 x 6.02 x 10^23 = 9.632x10^23|||||1. 2Cu (s) + O2 (g) -->> 2CuO (s) a. relative mass of Cu = 63.5 number of mole of Cu burnt = 5 / 63.5 = 0.07874mol By equation, number of mole of CuO formed = 0.7874mol relative mass of CuO = 63.5+18 = 81.5 mass of CuO = 81.5 x 0.07874 = 6.42 gram
F.4 CHEM
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發問:1. 5.0g of copper is burnt in air.a)find the mass of oxide formed.b)the experimental mass of oxide formed is usually less than the theoretical answer in(a).explain why.2. 24.6g of calcium was added into a beaker of water. Determine the mass of gas evolved.3. determine the number of atoms in the... 顯示更多 1. 5.0g of copper is burnt in air. a)find the mass of oxide formed. b)the experimental mass of oxide formed is usually less than the theoretical answer in(a).explain why. 2. 24.6g of calcium was added into a beaker of water. Determine the mass of gas evolved. 3. determine the number of atoms in the following compound. a)64g of argon gas help me please!!thanks..
最佳解答:
1. a) 2Cu + O2 → 2CuO Molar ratio Cu : CuO = 2 : 2 = 1 : 1 Molar mass of Cu = 63.5 g/mol Molar mass of CuO = 63.5 + 16 = 79.5 No. of moles of Cu reacted = 5/63.5 = 0.0787 mol No. of moles of CuO formed = 0.0787 mol Mass of CuO formed = 0.0787 * 79.5 = 6.26 g ...... (Answer) b) The copper contains some impurities which does not reacted. Or: The copper is not completely reacted. 2. Ca + 2H2O → Ca(OH)2 + H2 Mole ratio Ca : H2 = 1 : 1 Molar mass of Ca = 40 g/mol Molar mass of H2 = 1*2 = 2 g/mol No. of moles of Ca used = 24.6/40 = 0.615 mol No. of moles of H2 formed = 0.615 mol Mass of H2 formed = 0.615 * 2 = 1.23 g ...... (Answer) 3. Molar mass of Ar = 40 g/mol No. of moles of Ar = 64/40 = 1.6 mol No. of Ar atoms = 1.6 * (6.02 * 1023) = 9.63 * 1023 ...... (Answer)
其他解答:
1a. 2Cu(s) + O2(g) --> 2CuO(s) mole of Cu = 5/63.5 from the equation, 2 mole Cu produced 2 mole CuO then 5/63.5 mole Cu produced 5/63.5 mole CuO therefore, mass of oxide formed = 5/63.5 x (63.5+16) = 6.26g 1b. There may have loss of material during the experiment 2. Ca + H2O --> CaO + H2 mole of Ca used = 24.6/40.1 from the equation, 1 mole Ca form 1 mole H2 24.6/40.1 mole Ca form 24.6/40.1 mole H2 therefore, mass of H2 evolved = 24.6/40.1 x (1x2) = 1.23g 3. no.of mole of 64g Ar = 64/40 = 1.6 By the Avogadro Constant, 1 mole contains 6.02 x 10^23 atom therefore, no. of atoms that Ar contained = 1.6 x 6.02 x 10^23 = 9.632x10^23|||||1. 2Cu (s) + O2 (g) -->> 2CuO (s) a. relative mass of Cu = 63.5 number of mole of Cu burnt = 5 / 63.5 = 0.07874mol By equation, number of mole of CuO formed = 0.7874mol relative mass of CuO = 63.5+18 = 81.5 mass of CuO = 81.5 x 0.07874 = 6.42 gram
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