<span>массу железа, образующуюся при восстановлении</span>
Ответ:
Объяснение:
Mr(H3PO4) = 1*3 + 31 + 16*4 = 98
W(H) = Ar(H) *n / Mr(H3PO4) *100% = 1*3 / 98 *100% = 3%
W(P)= Ar(P) *n / Mr(H3PO4) *100% = 31*1 / 98 *100% = 32%
W(O) = Ar(O) *n / Mr(H3PO4) *100% = 16*4 / 98 *100% = 65%
Ca(OH)2+CO2=CaCO3+H2O
1 1
кол-во в-ва осадка СаСО3=0,4:100=0,004моль; объем СО2=0,004Х22,4=0,0896л
объемная доля СО2 в воздухе=(0,0896:112)Х100%=0,08%
На фотографии снизу все прописано.
1
[H+] * [OH-]= <span><span /></span><span><span><span><span /></span></span></span><span><span><span><span /></span></span></span><span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span></span>10^ -14
[H+] = <span><span><span /></span></span><span><span><span><span><span /></span></span></span></span><span><span><span><span><span /></span></span></span></span><span><span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span></span></span><span>10^ -14 / </span><span>10^ -9 =</span>10^ -5 <span>моль/л СРЕДА СЛАБОКИСЛАЯ</span>
2/
[OH-] = <span><span /></span><span><span><span><span /></span></span></span><span><span><span><span /></span></span></span><span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span></span>10^ -14 / 10^ -2 =10^ -12 моль/л СРЕДА <span>КИСЛАЯ
метилоранж красный
</span>
3.
[H+] = <span /><span><span><span /></span></span><span><span><span /></span></span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span>10^ -14 / 10^ -5 =10^ -9 моль/л
рН = -lg[H+] = 9 среда щелочная
лакмус - синий