А)1)2NaCl+500-600градув -> 2Na + Cl2<span>
2)2Na+2HOH=2NaOH+H2
3)2NaOH+H2SO4=Na2SO4+2HOH
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<span>CuO+<span>H2SO4=<span>CuSо4+<span>H2</span><span>O</span></span></span></span>
<span><span><span><span>n(Cuo)=m/M=20/80=0.25</span></span></span></span>
<span><span><span><span>n(Cuo)/1=n(<span>CuSo4)/1</span></span></span></span></span>
<span><span><span><span><span>n(CuSo4)=0.25</span></span></span></span></span>
<span><span><span><span><span>m<span>(CuSo4)=n*M=0.25*160=40г</span></span></span></span></span></span>
Ответ:
Объяснение:
На внешнем энергетическом уровне 2 электрона у: бериллий, галлий, радий