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NeyDa4nikn
@NeyDa4nikn
October 2021
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Химия. 8 класс. Решите, умоляю!
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2Al + 3S = Al₂S₃
4B + 3O₂ = 2B₂O₃
3) 2NaNO₃ = 2NaNO₂ + O₂
4) V₂O₅ + 5H₂ = 2V + 5H₂O
5) Al₂(SO₄)₃ + 3BaCl₂ = 3BaSO₄+ 2AlCl₃
6) 2Mg + O₂ = 2MgO
7) 2HgO = 2Hg + O₂
8) 3Mg + Fe₂O₃ = 3MgO + 2Fe
задача:
Дано:
m(PCl₅) = 10,85 г
Найти: m(Cl₂), V(Cl₂)
Решение:
Mr(PCl₅) = 31 + 35,5*5 = 208,5
Mr(Cl₂) 35,5*2 = 71
х л 10,85 г
2P + 5Cl₂ = 2PCl₅
112 л 417 г
х = 112 л *10,85 г\417 г = 2,9 л
n(Cl₂) = V\Vm *M = 2,9 л\22,4л\моль *71г\моль = 9,2 г
Ответ: m(Cl₂) = 9,2 г, V(Cl₂) = 2,9 л
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Answers & Comments
Verified answer
2Al + 3S = Al₂S₃4B + 3O₂ = 2B₂O₃
3) 2NaNO₃ = 2NaNO₂ + O₂
4) V₂O₅ + 5H₂ = 2V + 5H₂O
5) Al₂(SO₄)₃ + 3BaCl₂ = 3BaSO₄+ 2AlCl₃
6) 2Mg + O₂ = 2MgO
7) 2HgO = 2Hg + O₂
8) 3Mg + Fe₂O₃ = 3MgO + 2Fe
задача:
Дано:
m(PCl₅) = 10,85 г
Найти: m(Cl₂), V(Cl₂)
Решение:
Mr(PCl₅) = 31 + 35,5*5 = 208,5
Mr(Cl₂) 35,5*2 = 71
х л 10,85 г
2P + 5Cl₂ = 2PCl₅
112 л 417 г
х = 112 л *10,85 г\417 г = 2,9 л
n(Cl₂) = V\Vm *M = 2,9 л\22,4л\моль *71г\моль = 9,2 г
Ответ: m(Cl₂) = 9,2 г, V(Cl₂) = 2,9 л