Предмет: Химия,
автор: svetaismailova
расчитать относительные молекулярные массы веществ и массовые доли элементов в них если формулы этих веществ:HBr,K2CO3,Mg(OH)2,P2O5,FeCl3,Cu(NO3)2
Ответы
Автор ответа:
0
w=m(в-ва)m(р-ра)
1. M(HBr) = 1+80 = 81
w(H) = 181 = 0.01253*100% = 1.23%
w(Br) = 100%-1.23% = 98.77%
2. M(K₂CO₃) = 2*40+16*3+12 = 140
w(K₂) = 80140 = 0.57*100%= 57%
w(C) = 12140 = 0.086*100% = 8.6%
w(O₃) = 100%-57%-8.6% = 34.4%
3. M(Mg(OH)₂) = 24+16*2+1*2 = 58
w(Mg) = 2458 = 0.414*100% = 41.4%
w(O) = 3258 = 0.552*100% = 55.2%0
w(H) = 100%-41.4%-52.2% = 6.4%
4. M(P₂O₅) = 31*2+16*5 = 142
w(P₂) = 62142 = 0.423*100% = 42.3%
w(O₅) = 100%-42.3% = 57.7%
5. M(FeCl₃) = 56+35.5*3 = 162.5
w(Fe) = 56162,5 = 0.345*100% = 34.5%
w(Cl₃) = 100%-34.5% = 65.5%
6. M(Cu(NO₃)₂) = 64+28+96 = 188
w(Cu) = 64188 = 0.34*100% = 34%
w(N) = 28188 = 0.15*100% = 15%
w(O₃) = 100%-34%-15% = 51%
1. M(HBr) = 1+80 = 81
w(H) = 181 = 0.01253*100% = 1.23%
w(Br) = 100%-1.23% = 98.77%
2. M(K₂CO₃) = 2*40+16*3+12 = 140
w(K₂) = 80140 = 0.57*100%= 57%
w(C) = 12140 = 0.086*100% = 8.6%
w(O₃) = 100%-57%-8.6% = 34.4%
3. M(Mg(OH)₂) = 24+16*2+1*2 = 58
w(Mg) = 2458 = 0.414*100% = 41.4%
w(O) = 3258 = 0.552*100% = 55.2%0
w(H) = 100%-41.4%-52.2% = 6.4%
4. M(P₂O₅) = 31*2+16*5 = 142
w(P₂) = 62142 = 0.423*100% = 42.3%
w(O₅) = 100%-42.3% = 57.7%
5. M(FeCl₃) = 56+35.5*3 = 162.5
w(Fe) = 56162,5 = 0.345*100% = 34.5%
w(Cl₃) = 100%-34.5% = 65.5%
6. M(Cu(NO₃)₂) = 64+28+96 = 188
w(Cu) = 64188 = 0.34*100% = 34%
w(N) = 28188 = 0.15*100% = 15%
w(O₃) = 100%-34%-15% = 51%
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