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Predator231
@Predator231
July 2022
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какую массу соли можно получить при взаимодействии серной кислоты с магнием 9.6 г
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kopatel228
Mg + H2SO4 => MgSO4 + H2
n(Mg) = n(MgSO4)
n(Mg) = m/M = 9.6 / 24 = 0.4 моль
m(MgSO4) = n * M = 0.4 * (24 + 32 + 64) = 0.4 * 120 = 48 г
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Answers & Comments
n(Mg) = n(MgSO4)
n(Mg) = m/M = 9.6 / 24 = 0.4 моль
m(MgSO4) = n * M = 0.4 * (24 + 32 + 64) = 0.4 * 120 = 48 г