Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46
Depression is freezing point is a colligative property. It is mathematically expressed as ΔTf = Kf X m
where Kf = <span>freezing point depression constant = 1.86°c kg /mol (for water)
m = molality of solution = 1.40 m
</span>∴ ΔTf = Kf X m = 1.86 X 1.40 = 2.604 oC
Now, for water freezing point = 0 oC
∴Freezing point of solution = -2.604 oC
If, (.525m)(100 cm/m)(1/2.54 in/cm) = 20.7 in or C
Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
The ionic equation is as below
Ca^2+(aq) + SO4^2-(aq) ---> CaSO4(s)
EXPLANATION
K2SO4(aq) +Cai2(aq) ---> CaSO4(s) + Ki (aq)
ionic equation
= 2K^+(aq) + SO4^2-(aq) + Ca^2+(aq) + 2i^-(aq) --->CaSO4(s) + 2K^+(aq) +2 i^-(aq)
cancel the spectator ions that is 2k^+ and 2i^-
The net ionic equation is therefore
= Ca^2+(aq) + SO4^2-(aq) ----> CaSO4(s)
