0,2 x
H2SO4+K2CO3=K2SO4+CO2+H2O
1 1
m(K2CO3)=m(р-ра)хω=110х0,25=27,5 г.
n(K2CO3)=m/M=27,5÷138=0,2 моль
n(CO2)=x=0,2х1÷1=0,2моль
V(CO2)=Vmхn=22,4л./моль×0,2моль=4,48л.
ответ: 4,48л.
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Ряд элементов расположенных в порядке уменьшения радиусов атомов: 1) as, se, br. 2)ca, si, mg. 3) n, c, b. 4) be, li, na.