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auzdenov48
@auzdenov48
July 2022
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Вычислите относительную молекулярную массу и массовую долю кислорода в молекулах: CuO, Al(OH)3, NH4OH
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Влад1488228
CuO = 64+16 = 80
Cu = 80%
O = 20%.
Al(OH)3 = 27 + 16*3 + 3 = 78
Al = 34%
O = 62%
H = 4%.
NH4OH = 14 + 5 + 16 = 35
N = 40%
H = 14%
O = 46%
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Answers & Comments
Cu = 80%
O = 20%.
Al(OH)3 = 27 + 16*3 + 3 = 78
Al = 34%
O = 62%
H = 4%.
NH4OH = 14 + 5 + 16 = 35
N = 40%
H = 14%
O = 46%