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ответ:1. а) CuO- оксид меди(II) , б) N2O3-оксид азота(III), в) СаO-окисид кальция.
2. а) 4P+ 5O2 =2P205-сумма коэффициентов= 11
б) 2Cr(OH)3= Cr2O3 + 3H2Oсумма коэффициентов=6
3. Mr(Fe(OH)2=90 w =56/90*100%=62%
Mr(Fe2O3)=56*2+16*3=160 w =56*2/160*100%=70%
массовая доля больше в Fe2O3
Объяснение: