The freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
Using the equation,
Δ
= i
m
where:
Δ = change in freezing point (unknown)
i = Van't Hoff factor
= freezing point depression constant
m = molal concentration of the solution
Molality is expressed as the number of moles of the solute per kilogram of the solvent.
Molal concentration is as follows;
MM KCl = 74.55 g/mol
molal concentration = 
molal concentration = 0.1219m
Now, putting in the values to the equtaion Δ = im we get,
Δ = 2 × 1.86 × 0.1219
Δ = 0.4536°C
So, Δ of solution is,
Δ
= 0.00°C - 0.45°C
Δ = - 0.45°C
Therefore,freezing point of a solution containing 5. 0 grams of KCl and 550.0 grams of water is - 0.45°C
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Answer:
Explanation:
The breakdown reaction of ozone is as follows
It can be seen that 2 moles of ozone is required in the complete cycle
So for 10 cycles, 20 moles of ozone is required
m = Mass of 
= 15.5 g
M = Molar mass of = 104.46 g/mol
P = Pressure = 24.5 mmHg
T = Temperature = 232 K
R = Gas constant = 
Number of moles is given by
From ideal gas law we have
For 20 cycles of the reaction the volume of the ozone is .
Groups IVB–VIII, IB, and IIB, or 4–12
Answer:
Explanation — This page looks at the oxidation of alcohols using acidified sodium or ... of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. ... The electron-half-equation for this reaction is as follows: ... To do that, oxygen from an oxidizing agent is represented as [O]. ... Article type: Section or Page.
