Answer:
6960 J/kg°C
Explanation:
specific heat= mass×specific heat capacity×increase in temperature
specific heat= 0.240×1450×20= 6960 J/kg°C
hope it helps!
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
= i
m 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:
AM
Explanation:
to go from moles to grams you multiply by the Atomic Mass or Molar Mass (Atomic Mass for an element and Molar Mass for a compound).
The formula is:
Mass = moles * MM
Answer:
95.7 g CO to the nearest tenth.
Explanation:
2C + O2 ---> 2CO
Using relative atomic masses:
24 g C produces 2*12 + 2*16 g CO.
So 41 g produces ( (2*12 + 2*16) * 41 ) / 24
= 95.7 g CO,