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1mikro
@1mikro
August 2022
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определите массовую долю каждого элемента в йодиде магния
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krioira
Mr(MgI2) = Ar(Mg) + Ar(I)*2 = 24 + 127*2 = 278
W(Mg) = Ar(Mg) * n / Mr(MgI2) *100% = 24 *1 / 278 *100% = 8,6%
W(I) = Ar(I) *2 / Mr(MgI2) *100% = 127 *2 / 278 *100% = 91,4%
1 votes
Thanks 1
svetka1574
Verified answer
MgI2
Mr(MgI2)=24+2*127=278
W(Mg)=24/278=0.09=9%
W(I)=254/278=0.91=91%
1 votes
Thanks 2
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Answers & Comments
W(Mg) = Ar(Mg) * n / Mr(MgI2) *100% = 24 *1 / 278 *100% = 8,6%
W(I) = Ar(I) *2 / Mr(MgI2) *100% = 127 *2 / 278 *100% = 91,4%
Verified answer
MgI2Mr(MgI2)=24+2*127=278
W(Mg)=24/278=0.09=9%
W(I)=254/278=0.91=91%