We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
The last row going across
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
Answer:
wHeRe ArE tHe StAtEmEnTs mAn
Answer:
Explanation:
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