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vladanesterchuk
@vladanesterchuk
June 2022
2
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Обчисліть об’єм газу (н. у.), який виділиться під час електролізу водного розчину плюмбум(ІІ) нітрату масою 120 г з масовою часткою розчиненої речовини 7,5 %.
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dec77cat
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0 votes
Thanks 1
mrvladimir2
Verified answer
Pb(NO₃)₂ = Pb²⁺ + 2NO₃⁻
(K-) Pb²⁺ + 2e = Pb 2
(A+) 2H₂O - 4e = O₂ + 4H⁺ 1
2Pb(NO₃)₂ + 2H₂O = 2Pb + O₂ + 4HNO₃
m(Pb(NO₃)₂)=mw
m(Pb(NO₃)₂)/[2M(Pb(NO₃)₂] = V(O₂)/V₀
mw/[2M(Pb(NO₃)₂] = V(O₂)/V₀
V(O₂)=V₀mw/[2M(Pb(NO₃)₂]
V(O₂)=22,4л/моль*120г*0,075/[2*331,2г/моль] =
0,30
л
1 votes
Thanks 0
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Answers & Comments
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Verified answer
Pb(NO₃)₂ = Pb²⁺ + 2NO₃⁻(K-) Pb²⁺ + 2e = Pb 2
(A+) 2H₂O - 4e = O₂ + 4H⁺ 1
2Pb(NO₃)₂ + 2H₂O = 2Pb + O₂ + 4HNO₃
m(Pb(NO₃)₂)=mw
m(Pb(NO₃)₂)/[2M(Pb(NO₃)₂] = V(O₂)/V₀
mw/[2M(Pb(NO₃)₂] = V(O₂)/V₀
V(O₂)=V₀mw/[2M(Pb(NO₃)₂]
V(O₂)=22,4л/моль*120г*0,075/[2*331,2г/моль] = 0,30 л