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i19762007
@i19762007
July 2022
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расчитать массовою долю серы 1 соединениях SO2,AL2S3,AL2(SO4)3
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DJToxa
Verified answer
1.
M(SO2)=64г/моль
w(S)=32/64=0.5=50%
2.
M(Al2S3)=54+96=150г/моль
w(S)=96/150=0.64=64%
3.
M(Al2(SO4)3)=54+96+196=346г/моль
w(S)=96/342=0.28=28%
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Answers & Comments
Verified answer
1.M(SO2)=64г/моль
w(S)=32/64=0.5=50%
2.
M(Al2S3)=54+96=150г/моль
w(S)=96/150=0.64=64%
3.
M(Al2(SO4)3)=54+96+196=346г/моль
w(S)=96/342=0.28=28%