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omeli [17]
3 years ago
15

The gas with an initial volume of 24.0 L at a pressure of 565 torr is compressed until the volume is 16.0 L. What is the final p

ressure of the gas, assuming the temperature and amount of gas does not change
Chemistry
1 answer:
Reil [10]3 years ago
8 0

Answer:

848 torr  

Explanation:

The only variables are the pressure and the volume, so we can use Boyle's Law.

p₁V₁ = p₂V₂

Data:

p₁ = 565 torr; V₁ = 24.0 L

p₂ = ?;            V₂ =  16.0 L

Calculations:

\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}

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alexandr1967 [171]

Answer:

236.9g

Explanation:

Given parameters:

Volume of gas  = 81.3L

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temperature of gas = 95°C

Unknown:

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Mathematically, it is expressed as;

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the unknown here is n which is the number of moles;

P is the pressure, V is the volume, R is the gas constant and T is the temperature.

convert pressure into atm

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Convert temperature to Kelvin;   95 + 273  = 368K

       2 x 81.3 = n x 0.082 x 368

             n = \frac{2 x 81.3}{0.082 x 368}   = 5.38moles

Since the unknown is mass;

 Mass  = number of moles x molar mass

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5 0
3 years ago
At what temperature will 0.100 molal (M) NaCl(aq) boil? <br> Kb= 0.51 C kg mol^-1
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The increase of the boling point of a solution is a colligative property.

The formula for the increase of the normal boiling point of water is:

ΔTb = Kb * m

Where m is the molallity of the solution and Kb is the molal boiling constant in °C/mol.

ΔTb = 0.51 °C / m * 0.100 m = 0.051 °C.

So, the new boiling temperature is Tb = 100°C + 0.051°C = 100.051 °C.

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Hence, the correct answer is Option B.

4 0
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