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alinanurtazina11
@alinanurtazina11
July 2022
1
9
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найти массовую долю всех элементов
KMnO₄
и H₃PO₄
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liana5105
1)Mr(KMnO4)=39+55+16×4=94+64=158г/моль
w(K)=Ar(K)/Mr(KMnO4)×100%=39/158=0,25×100%=25%
w(Mn)=Ar(Mn)/Mr(KMnO4)×100%=55/158=0,35×100%=35%
w(O)=Ar(O)/Mr(KMnO4)×100%=0,40×100%=40%
2)Mr(H3PO4)=1×3+31+16×4=34+64=98г/моль
w(H)=Ar(H)/Mr(H3PO4)×100%=3/98=0,03×100%=3%
w(P)=Ar(P)/Mr(H3PO4)×100%=31/98=0,32×100%=32%
w(O)=Ar(O)/Mr(H3PO4)×100%=64/98=0,65×100%=65%
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Answers & Comments
w(K)=Ar(K)/Mr(KMnO4)×100%=39/158=0,25×100%=25%
w(Mn)=Ar(Mn)/Mr(KMnO4)×100%=55/158=0,35×100%=35%
w(O)=Ar(O)/Mr(KMnO4)×100%=0,40×100%=40%
2)Mr(H3PO4)=1×3+31+16×4=34+64=98г/моль
w(H)=Ar(H)/Mr(H3PO4)×100%=3/98=0,03×100%=3%
w(P)=Ar(P)/Mr(H3PO4)×100%=31/98=0,32×100%=32%
w(O)=Ar(O)/Mr(H3PO4)×100%=64/98=0,65×100%=65%