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August 2021
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Ребята, надо найти массовую долю CrN(O3)3, BaCO3, Ag2CO3, MgIO3,NaPO3 (20 баллов ) heeeeeelp
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Alexei78
Cr(NO3)3
M(Cr(NO3)3)=52+(14+16*3)*3=238g/mol
W(Cr)=52/238*100%=21.85%
W(N)=14*3/238*100%=17.65%
W(O)=16*9/238*100%=60.5%
BaCO3
M(BaCO3)=137+12+16*3=197g/mol
W(Ba)=137/197*100%=69.54%
W(C)=12/197*100%=6.09%
W(O)=16*3/197*100%=24.37%
Ag2CO3
M(Ag2CO3)=108*2+12+16*3=276g/mol
W(Ag)=108*2/276*100%=76.26%
W(C)=12/276*100%=4.35%
W(O)=16*3/276*100%=17.39%
MgIO3
M(MgIO3)=24+127+16*3=199g/mol
W(Mg)=24/199*100%=12.06%
W(I)=127/199*100%=63.82%
W(O)=16*3/199*100%=24.12%
NaPO3
M(NaPO3)=23+31+16*3=102g/mol
W(Na)=23/102*100%=22.55%
W(P)=31/102*100%=30.39%
W(O)=16*3/102*100%=47.06%
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Answers & Comments
M(Cr(NO3)3)=52+(14+16*3)*3=238g/mol
W(Cr)=52/238*100%=21.85%
W(N)=14*3/238*100%=17.65%
W(O)=16*9/238*100%=60.5%
BaCO3
M(BaCO3)=137+12+16*3=197g/mol
W(Ba)=137/197*100%=69.54%
W(C)=12/197*100%=6.09%
W(O)=16*3/197*100%=24.37%
Ag2CO3
M(Ag2CO3)=108*2+12+16*3=276g/mol
W(Ag)=108*2/276*100%=76.26%
W(C)=12/276*100%=4.35%
W(O)=16*3/276*100%=17.39%
MgIO3
M(MgIO3)=24+127+16*3=199g/mol
W(Mg)=24/199*100%=12.06%
W(I)=127/199*100%=63.82%
W(O)=16*3/199*100%=24.12%
NaPO3
M(NaPO3)=23+31+16*3=102g/mol
W(Na)=23/102*100%=22.55%
W(P)=31/102*100%=30.39%
W(O)=16*3/102*100%=47.06%