Находимо кількість моль водню і кисню? n(H2)=10/2=5
n(O2)=44.8/22.4=2
відповідно в 10г (Н2) більше молекул
Константа диссоциации азотистой кислоты Кдисс = 4*10⁻⁴
Концентрация протонов водорода [H⁺] = √ Кдисс * См = √4*10⁻⁴*0,01 =2*10⁻³ моль/л
рН = -lg[H⁺] = -lg 2*10⁻³ = 2,7
Ответ: рН = 2,7
<span>фосфор <span>1s</span></span>² <span><span>2s</span></span>²<span><span> 2p</span></span>⁶<span><span> 3s</span></span>²<span><span> 3p</span></span>³<span><span><span>
</span></span></span><span>фтор <span>1s</span></span>² <span><span>2s</span></span>² <span><span>2p</span></span>⁵<span><span><span>
</span></span></span><span>аргон <span>1s</span></span>²<span><span> 2s</span></span>²<span><span> 2p</span></span>⁶<span><span> 3s</span></span>²<span><span> 3p</span></span>⁶<span><span><span>
</span></span></span><span>алюминий <span>1s</span></span>²<span><span> 2s</span></span>²<span><span> 2p</span></span>⁶<span><span> 3s</span></span>²<span><span> 3p</span></span>¹<span><span><span>
</span></span></span><span>сера <span>1s</span></span>²<span><span> 2s</span></span>²<span><span> 2p</span></span>⁶<span><span> 3s</span></span>²<span><span> 3p</span></span>⁴<span><span><span>
</span></span></span><span>калий <span>1s</span></span>²<span><span> 2s</span></span>²<span><span> 2p</span></span>⁶<span><span> 3s</span></span>²<span><span> 3p</span></span>⁶ <span><span>4s</span></span>¹
2Cu(NO3)2 = 2CuO + 4NO2 + O2
По уравнению реакции:
n(NO2)= 4 моль
V(NO2)=22,4*4=89,6 л
n(O2)=1 моль
V(O2)=1*22,4=22,4 л
Ответ:V(NO2)=89,6 л
V(O2)=22,4 л
8Al<span> + 15H2SO4(</span>конц., гор.)<span> → 4Al2(SO4)3 + 3H2S↑ + 12H2O
</span>
6НСl + 2Al = 2AlCl3 + 3H2,
3Н2SO4(разб<span>) + 2Al = Al2(SO4)3 + 3H2</span>