HCl + NaOH -> NaCl + H2O
reactants: hydrochloric acid HCl
sodium hydroxide NaOh
products: sodium chloride (table salt) NaCl
dihydrogen monoxide (water)
H2O
Answer:
Solution given:
1 mole of KCl
22.4l
1 mole of KCl74.55g
we have
0.14 mole of KCl74.55*0.14=10.347g
74.55g of KCl22.4l
10.347 g of KCl22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
Answer:
the answer is pH of 7 I don't no Hydrochloric hijinx work
Chemical reaction: C₃H₇COOH → C₃H₇COO⁻ + H⁺.
c(<span>butanoic acid) = 0,100 M.
</span>α = 1,23% = 0,0123.
Ka = α² · c / 1 - α.
Ka = 0,0123² · 0,1 M / 1 - 0,0123.
Ka = 0,0000153 M.
Ka = c(C₃H₇COO⁻) · c(H⁺) / c(C₃H₇COOH).
c(H⁺) = α · c(C₃H₇COOH).
c(H⁺) = 0,0123 · 0,1 M = 0,00123 mol/L.
pH = -log c(H⁺).
pH = 2,91.
