Answer:
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Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
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Answer:
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Explanation:
The equation is given as;
2HCl(aq) + Zn(s) + H2(g) + ZnCl2(aq)
In writing an ionic equation, only the aqueous compounds dissociates into ions. This means HCl and ZnCl2 would dissociate to form ions.
This is given as;
2H+ + 2Cl- + Zn(s) --> H2(g) + Zn2+ + 2Cl-
The correct option is;
b) 2H+(aq) + 2C1-(aq) + Zn(s) → H2(g) + Zn2+(aq) + 2Cl-(aq)
Answer: 2.78 moles of molecular oxygen will occupy 62.22 liters.
Explanation:
