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timosikkarina
@timosikkarina
August 2021
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обчисліть масові частки хімічних елементів (у відсотках) у формулі сполуки (nh4)2 [fe (so4)2]
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Alexei78
M(NH4)2{Fe(SO4)2} = (14+4)*2+{56+(32+16*4)*2}=284 g/mol
W(N)=14*2 / 284*100% =9.86%
W(H)=1*8 / 284*100% = 2.82 %
W(Fe) = 56/284*100% = 19.72%
W(S)=32*2/284*100% = 22.54 %
W(O)=16*8/284*100% = 45.07%
2 votes
Thanks 9
jnona
Mr(NH4)2{Fe(SO4)2}=284
ω(N)=2800/284=9,9%
ω(H)=800/284=2,9%
ω(Fe)=5600/284=19,7%
ω(S)=6400/284=22,5%
ω(O)=12800/284=45%
3 votes
Thanks 1
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Answers & Comments
W(N)=14*2 / 284*100% =9.86%
W(H)=1*8 / 284*100% = 2.82 %
W(Fe) = 56/284*100% = 19.72%
W(S)=32*2/284*100% = 22.54 %
W(O)=16*8/284*100% = 45.07%
ω(N)=2800/284=9,9%
ω(H)=800/284=2,9%
ω(Fe)=5600/284=19,7%
ω(S)=6400/284=22,5%
ω(O)=12800/284=45%