1)
2H2O=>2H2+O2
AgNo3+AlCl3=>AgCl3+Al(NO3)3
SO2<span> + </span>K2O<span> = </span>K2<span>SO3</span>
<span>Na2S<span> + </span>H2SO4<span>(разб.)=SO2 + </span>Na2SO4<span> + </span>H2<span>O</span></span>
<span><span>2)</span></span>
14.4:65=0.22моль Zn
V(H2)=22.4*0.22=5 л
3)
H2SO4<span>+</span>BaCl2<span>=BaSO4+2HCl</span>
<span>KNO3+NaoH=>KOH+NaNO3</span>
<span>H2+Cl2=>2HCl</span>
+1 -2 +1-2 +7 -2 +2 -2 +4 -2
H2O H2S Cl2O7 ZnS CO2
Найти CxHyNz,если дано C,H,N,D(возд.)=2,5,w(N)=19,18%,w(H)=15,06%,w(C)=65,75%
Матвей Переход [13]
Дано:
<span>w
(C) = 65,75%</span>
<span>w
(H) = 15,06%</span>
<span>w
(N) = 19,18%</span>
<span>D(возд)
= 2,5</span>
<span>Найти: CxHyNz</span>
Решение:
<span>Mr (в-ва) = 29 ∙
2,5 = 72,5</span>
<span>n (э) = w(э) ∙ Mr(в-ва) \Ar(э) ∙ 100%</span>
<span>n (C) =
65,75% ∙ 72,5 \12 ∙ 100% = 3,9 ≈ 4</span>
<span>n (H) =
15,06% ∙ 72,5\1 ∙<span> 100% = 10,9 </span>≈ 11</span>
<span>n (N) =
19,18% ∙ 72,5\14 ∙ 100% = 0,99 ≈ 1</span>
n (C) : n
(H) : n(N) = 4:11:1 = CxHyNz = C4H11N1
<span>ответ: C4H11N1 </span>
Na(+) + SO4(2-) + Ba(2+) + 2Cl(-) = 2Na(+) + 2CL(-) +BASO4
SO4(2-) + Ba(2+)=BASO4- выпадает в осадок
CuSO₄ + 2NaOH = Cu(OH)₂ + Na₂SO₄
1+2+1+1=5
D) 5