Дано
D(He) = 5.4
W(C) = 58.82% = 0.5882
W(H) = 9.8% = 0.098
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CxHyOz-?
M(CxHyOz) = 20*D(He)= 20*5.4 = 108 г/моль
W(O)= 100% - (W(C)+W(H)) = 100 - (58,82 + 9.8) = 31.38% = 0.3138
CxHyOz / 108 = 0.5882 / 12 : 0.098 / 1 : 0.3138 / 16
CxHyOz = 5 : 11 : 2
C5H11O2
C4H10COOH
ответ Бутановая кислота
XL 2mol
2NH3+H3PO4=(NH4)2HPO4
44.8L 1mol
x=44.8·2/1=89.6L
Mr(Al2(SO4)3= 27*2+32*3+16*12=54+96+192=342
Zn + H₂SO₄ = ZnSO₄ + H₂
Zn + 2H⁺+SO₄²⁻ = Zn²⁺ + SO₄²⁻ + H₂
Zn + 2H⁺ = Zn²⁺ + H₂
MgO + H₂SO₄ = MgSO₄ + H₂O
MgO + 2H⁺ + SO₄²⁻ = Mg²⁺ + SO₄²⁻ + H₂O
MgO + 2H⁺ = Mg²⁺ + H₂O
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
2Na⁺ + 2OH⁻ + 2H⁺ + SO₄²⁻ = 2Na⁺ + SO₄²⁻ + 2H₂O
2OH⁻ + 2H⁺ = 2H₂O
H⁺ + OH⁻ = H₂O