Q = m c T
c= 0.140 j/(g x °c)
m= 250.0g
T =52
hope you can solve it now
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
Learn more about freezing point here;
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Answer A )
Potassium chloride
K+ Cl-
hope this helps!.
Explanation:
It is often desirable to determine the mass percent of elements in a given compound.
To determine the mass percent of elements:
- Evaluate the formula mass of the compound. This is done by summing the atomic masses of the atoms in the compound together.
- The mass percentage is determined by pacing the mass contribution of each element or group to the formula mass of the compound and multiply by 100.
Learn more:
Percent by mass brainly.com/question/5544078
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