Answer:
Boiling point: 63.3°C
Freezing point: -66.2°C.
Explanation:
The boiling point of a solution increases regard to boiling point of the pure solvent. In the same way, freezing point decreases regard to pure solvent. The equations are:
<em>Boiling point increasing:</em>
ΔT = kb*m*i
<em>Freezing point depression:</em>
ΔT = kf*m*i
ΔT are the °C that change boiling or freezing point.
m is molality of the solution (moles / kg)
And i is Van't Hoff factor (1 for I₂ in chloroform)
Molality of 50.3g of I₂ in 350g of chloroform is:
50.3g * (1mol / 253.8g) = 0.198 moles in 350g = 0.350kg:
0.198 moles / 0.350kg = 0.566m
Replacing:
<em>Boiling point:</em>
ΔT = kb*m*i
ΔT = 3.63°C/m*0.566m*1
ΔT = 2.1°C
As boiling point of pure substance is 61.2°C, boiling point of the solution is:
61.2°C + 2.1°C = 63.3°C
<em>Freezing point:</em>
ΔT = kf*m*i
ΔT = 4.70°C/m*0.566m*1
ΔT = 2.7°C
As freezing point is -63.5°C, the freezing point of the solution is:
-63.5°C - 2.7°C = -66.2°C
When the value of Ksp = 3.83 x 10^-11 (should be given - missing in your Q)
So, according to the balanced equation of the reaction:
and by using ICE table:
Ag2CrO4(s) → 2Ag+ (Aq) + CrO4^2-(aq)
initial 0 0
change +2X +X
Equ 2X X
∴ Ksp = [Ag+]^2[CrO42-]
so by substitution:
∴ 3.83 x 10^-11 = (2X)^2* X
3.83 x 10^-11 = 4 X^3
∴X = 2.1 x 10^-4
∴[CrO42-] = X = 2.1 x 10^-4 M
[Ag+] = 2X = 2 * (2.1 x 10^-4)
= 4.2 x 10^-4 M
when we comparing with the actual concentration of [Ag+] and [CrO42-]
when moles Ag+ = molarity * volume
= 0.004 m * 0.005L
= 2 x 10^-5 moles
[Ag+] = moles / total volume
= 2 x 10^-5 / 0.01L
= 0.002 M
moles CrO42- = molarity * volume
= 0.0024 m * 0.005 L
= 1.2 x 10^-5 mol
∴[CrO42-] = moles / total volume
= (1.2 x 10^-5)mol / 0.01 L
= 0.0012 M
by comparing this values with the max concentration that is saturation in the solution
and when the 2 values of ions concentration are >>> than the max values o the concentrations that are will be saturated.
∴ the excess will precipitate out
Answer:
The new volume of gas is 25.7 L.
Explanation:
Given data:
Initial volume = 24.0 L
Initial pressure = 10.0 atm
Initial temperature = 200 K
Final temperature = 300 K
Final volume = ?
Final pressure = 14.0 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 10.0 atm × 24.0 L × 300K / 200 K × 14.0 atm
V₂ = 72000 atm .L. K / 2800 K.atm
V₂ = 25.7 L
The new volume of gas is 25.7 L.
Answer:
d
Explanation:
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