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pennyleon
@pennyleon
March 2022
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Даю 20 баллов
5 задание в контрольной по неметаллам
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gavrilka
NaCl + AgNO3 = AgCl + NaNO3
Cl(-) + Ag(+) = AgCl
m(NaCl)/M(NaCl)=m(AgCl)/M(AgCl) => m(AgCl)=m(NaCl)*M(AgCl)/M(NaCl)=5,85*143,5/58,5=14,35 г
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Answers & Comments
Cl(-) + Ag(+) = AgCl
m(NaCl)/M(NaCl)=m(AgCl)/M(AgCl) => m(AgCl)=m(NaCl)*M(AgCl)/M(NaCl)=5,85*143,5/58,5=14,35 г