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3 years ago

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If 2.40 moles of gas are held at a temperature of 97.0 °C in a container with a volume of 45.0 L, what is the pressure of the ga

s?

1 answer:

Rainbow [258]3 years ago

7 0

PV=nRT<=> P=nRT/V=2,40*R*(273+97)/45 atm.
Calculate it. R is a number that is given, find it and use your math to solve.

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Answer:

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A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b

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Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.

1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>

</span>

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3 years ago