1)AlBr3+HOH=AlOHBr2+HBr
Al(3+) +3Br(-) +HOH=AlOH(2+) +2Br(-) +H(+) +Br(-)
<span>Al(3+) +HOH=AlOH(2+) +H(+)
2)</span>FeCl2<span> + HOH ↔ FeOHCl + HCl</span>
Fe2+<span> + 2Cl</span>-<span> + H</span>2O ↔ FeOH+<span> + Cl</span>-<span> + H</span>+<span> + Cl</span>-
Fe2+<span> + H</span>2O ↔ FeOH+<span> + H</span><span>+
3)</span> CuSO4<span> + 2H</span>2O = Cu(OH)2<span> + H</span>2SO4
Cu2+<span> + SO</span>42-<span> + 2H</span>2O = Cu(OH)2<span> + 2H</span>+<span> + SO</span>4<span>2-
4)</span><span>KNO2 + H2O = KOH + HNO2 </span>
<span>K+ + NO2- + H2O = K+ + OH- + HNO2 </span>
<span>NO2- + H2O = OH- + HNO2</span><span>
</span>
HCOH+O2=CO2+H2O.................................................
Ответ:
7,638 л
Объяснение:
M(CO2)=12+16*2=12+32=44 г/моль
n(CO2)=m(CO2)/M(CO2)=15/44=0.341
Vн.у=22,4 л/моль
V(CO2)=0.341*22.4=7.638 л
CaCO3*MgCO3 = CaO + MgO + 2CO2↑
<span>n(CO2) = V(CO2)/Vm=21/22,4=0,94 моль
</span><span>По уравнению n(CaCO3*MgCO3) = n(CO2)/2 = 0,94/2=0,47 моль
</span><span>m(CaCO3*MgCO3) = n(CaCO3*MgCO3)*M(CaCO3*MgCO3) =0,47*184=86,48 г
</span><span>m(примесей) = m(доломита) - m(CaCO3*MgCO3) = 100-86,48=13,52 г
</span><span>w(примесей) = m(примесей)/m(доломита) =13,52/100 *100%=13,52%
</span>Ответ:13,52%
N(N2)=5.6/28=0.2 моль
N=6*10^23*0.2=12*10^22 молекул
