Вычислить массовую долю Хлорк в хлоропреновом каучуке

Ответ:

Mr(NO2)=14+2*16=46

W(N)=14/46=0.304=30.4%

W(O)=32/46=0.696=69.6%

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Mr(PbCl2)=207+2*35.5=278

W(Pb)=207/278=0.745=74.5%

W(Cl)=71/278=0.255=25.5%

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Mr(Na2CO3)=2*23+12+3*16=106

W(Na)=46/106=0.434=43.4%

W(C)=12/106=0.113=11.3%

W(O)=48/106=0.453=45.3%

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Mr(H2SO4)=2*1+32+4*16=98

W(H)=2/98=0.020=2%

W(S)=32/98=0.327=32.7%

W(O)=64/98=0.653=65.3%

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Mr(Ba3(PO4)2)=3*137+2*31+8*16=601

W(Ba)=411/601=0.684=68.4%

W(P)=62/601=0.103=10.3%

W(O)=128/601=0.213=21.3%

Ответ:

Объяснение:

Mr(NO2)=14+2*16=46

W(N)=14/46=0.304=30.4%

W(O)=32/46=0.696=69.6%

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Mr(PbCl2)=207+2*35.5=278

W(Pb)=207/278=0.745=74.5%

W(Cl)=71/278=0.255=25.5%

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Mr(Na2CO3)=2*23+12+3*16=106

W(Na)=46/106=0.434=43.4%

W(C)=12/106=0.113=11.3%

W(O)=48/106=0.453=45.3%

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Mr(H2SO4)=2*1+32+4*16=98

W(H)=2/98=0.020=2%

W(S)=32/98=0.327=32.7%

W(O)=64/98=0.653=65.3%

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Mr(Ba3(PO4)2)=3*137+2*31+8*16=601

W(Ba)=411/601=0.684=68.4%

W(P)=62/601=0.103=10.3%

W(O)=128/601=0.213=21.3%

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