C17H35COOK + H2O = C17H35COOH + KOH
3Zn(OH)2 + 2H3PO4 = Zn3(PO4)2 + 6H2O
<span><span>Al<span>(OH)</span>2</span> → <span>AlO</span>+<span>H2O
</span></span><span><span>Fe<span>(OH)</span>2</span> → <span>H2O</span>+<span>FeO
</span></span><span><span>2<span>Fe<span>(OH)</span>3</span></span> → <span>Fe2O3</span>+<span>3<span>H2<span>O
</span></span></span></span>
Ответ:
дано
m(CaCO3) = 4 g
---------------------
m(Ca3(PO4)2)-?
3CaCO3+2H3PO4-->Ca3(PO4)2+3H2O+3CO2
M(CaCO3) = 100g/mol
n(CaCO3) = m/M = 4 / 100 = 0.04 mol
3n(CaCO3) = n(Ca3(PO4)2)
n(Ca3(PO4)2 = 0.04 / 3 = 0.013 mol
M(Ca3(PO4)2) = 310 g/mol
m(Ca3(PO4)2) = n*M = 0.013 * 310 = 4.03 g
ответ 4.03 г
Объяснение:
C₀=2 моль/л
v₀-?
v=0,5 л
c=0,05 моль/л
количество вещества кислоты
n=c₀v₀=cv
объем исходного раствора
v₀=cv/c₀
v₀=0,05моль/л*0,5л/2моль/л=0,0125 л = 12,5 мл