1) V(NH3)=n*Vm=0,5 моль*22,4дм^3/моль=11,2 дм^3
2) n(N2)=V/Vm=6,72л/22,4л/моль=0,3 моль
3) V(O2)=n*Vm=m/M*Vm=4г/32г/моль*22,4дм^3/моль=2,8 дм^3
4) m(N2)=n*M=V/Vm*M=4,48л/22,4дм^3/моль*28г/моль=5,6 г
<span>1. Zn+<span>2<span>HCl</span></span> → <span>ZnCl2</span>+<span>H2
</span></span><span><span>3<span>Li2O</span></span>+<span>2<span>H3PO4</span></span> → <span>2<span>Li3PO4</span></span>+<span>3<span>H2O
</span></span></span><span><span>CuCl2</span>+<span>2<span>KOH</span></span> → <span>Cu<span>(OH)</span>2</span>+<span>2<span>K<span>Cl
2. </span></span></span></span>Zn+H2SO4 → ZnSO4+H2
<span>Zn+<span>CuO</span> → <span>ZnO</span>+Cu
</span><span>Zn+<span>H2O</span> → <span>ZnO</span>+<span>H2
</span></span><span><span>H2SO4</span>+<span>CuO</span> →<span>CuSO4</span>+<span>H2O
</span></span><span><span>H2SO4</span>+<span>Ca<span>(OH)</span>2</span>→<span>CaSO4</span>+<span>2<span>H2O </span></span></span>
<span><span>H2SO4</span>+<span>CO2</span> →<span>H2CO3</span>+<span>SO3
</span></span><span><span>H2SO4</span>+<span>Na2O</span>→ <span>H2O</span>+<span>Na2SO4
</span></span><span><span>CuO</span>+<span>H2O</span> →<span>Cu<span>(OH)</span>2 </span></span>
<span><span>CuO</span>+<span>CO2</span> →<span>CuCO3
</span></span><span><span>H2O</span>+<span>Na2O</span> →<span>2<span>NaO<span>H </span></span></span></span>
Снизу написанно аммиачный р-р
Дано:
V(H2S) = 0.448 л
m(раствора K2S) = 110г
Найти: W(соли) -?
Решение:
2KOH+H2S = K2S + 2H20
n=V/Vm
n(H2S) = 0.448\22,4=0,02 моль
n(H2S) = n(K2S) = 1:1
n(K2S) = 0.02 моль
m(K2S) = 0.02*110 =2.2 г
W= m(чист вва)\m(р-ра) = 2,2/110= 0,02 = 2%
Ответ: 2%
1)Одинаковые: Нет
2)HNO3(63) , H2CO3(62)
3)HNO3(63), H3PO4(98)
