Two samples of sodium chloride were decomposed into their constituent elements. one sample produced 4.65 g of sodium and 7.16 g of chlorine, and the other sample produced 7.45 g of sodium and 11.5 g of chlorine. are these results consistent with the law of constant composition? explain your answer.
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Ответ:
1) m(Na) = 4.65 g.
n(Na) = m(NA) ÷ M(Na).
n(Na) = 4.65 g ÷ 23 g/mol.
n(Na) = 0.202 mol.
m(Cl₂) = 7.16 g.
n(Cl₂) = 7.16 g ÷ 70.9 g/mol.
n(Cl₂) = 0.101 mol → n(Cl) = 2 · n(Cl₂) = 0.202 mol.
In sodium chloride (NaCl) proportion of sodium and chlorine atoms are:
n(Na) : n(Cl) = 1 : 1.
In this example n(Na) : n(Cl) = 0.202 mol : 0.202 mol (1 : 1).
2) n(Na) = 7.45 g ÷ 23 g/mol.
n(Na) = 0.324 mol.
n(Cl₂) = 11.5 g ÷ 70.9 g/mol.
n(Cl₂) = 0.162mol → n(Cl) = 0.324 mol.
n(Na) : n(Cl) = 1 : 1; results consistent with the law of constant composition.
Ответ:
C-12:6 electrons
C-13:6 electrons
C-14: 6 electrons
Hope this helps ;w;