Answer:
0.15M
Explanation:
NaOH + HCl -->NaCl + H2O
moles of NaOH reacted with HCl
(0.5 × 30)/1000 = 1.5 × 10^ -3
so the moles of HCl reacted with NaOH
1.5 × 10^ -3
concentration of HCl = ( 1.5 × 10^ -3)/10 ×1000
Answer is: adding NaCl will lower the freezing point of a solution.
A solution (in this example solution of sodium chloride) freezes at a lower temperature than does the pure solvent (deionized water).
The higher the solute concentration (sodium chloride), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
Dissociation of sodium chloride in water: NaCl(aq) → Na⁺(aq) + Cl⁻(aq).
Answer:
C₅H₈O₂
Explanation:
methyl methacrylate = 100 amu
6.91g CO₂ = 0.157 moles
2.26g H₂O = 0.125 moles
0.157 ÷ 0.125 = 1.256
{(CO₂)₁.₂₅₆ + (H₂O)₁} × 4 = (CO₂)₅ + (H₂O)₄
C₅H₈O?
C₅ = 60.05 amu H₈ = 8.064 amu
60.05 + 8.064 = 68.114 amu
100 amu - 68.114 amu = 31.886 amu
O = 16 amu
O = 2