Mr(Cu2O)=Ar(Cu)*2+Ar(O)=64*2+16=144
w%(O)=16/144*100%=11,1%
б)Mr(Na2O)=23*2+16=62
w%(O)=16/62*100%=25,8%
в)Mr(Na2O2)=23*2+16*2=78
w%(O)=32/78*100%=41%
г)Mr(H2O)=1*2+16=18
w%(O)=16/18*100%=88,9%
ответ:наибольшая массовая доля кислорода в воде
2H2SO4 + Cu = CuSO4 + SO2 + 2Н2O<span>
H2SO4 + Ba(NO3)2 = </span><span><span>BaSO4</span> + <span>2<span>HNO<span>3
</span></span></span></span><span>H2SO4 + 2NaOH = </span><span><span>2<span>H2O</span></span> + <span>Na2SO<span>4
</span></span></span>H2SO4 + 2K = <span><span>K2SO4</span> + <span>H2</span></span>