Хлор
зраяд атома+17(по порядковому номеру елемента)
1) Дано
m(CH3OH) = 16 g
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m(CH3COOCH3) = ?
CH3COOH+CH3OH -->CH3COOCH3+H2O
M(CH3OH) = 32 g/mol
n(CH3OH) = m/M = 16 / 32 = 0.5 mol
n(CH3OH) = n(CH3COOCH3) = 0.5 mol
M(CH3COOCH3) = 44 g/mol
m(CH3COOCH3) = n*M = 0.5 * 44 = 22 g
ответ 22 г
2)
С2H6-->C2H4+H2
C2H4+HOH-->C2H5OH
C2H5OH-->CH3COH+H2
CH3COH+Ag2O-->CH3COOH+2Ag
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
а) бор B: 1.794*10^-23*6.02*10^23 = 10.79988 г/моль;
Na2O + 2HCl → 2NaCl + H2O