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zlopas [31]
3 years ago
14

A 0.245-L flask contains 0.467 mol co2 at 159 °c. Calculate the pressure using the ideal gas law.

Chemistry
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

Pressure, P = 67.57 atm

Explanation:

<u>Given the following data;</u>

  • Volume = 0.245 L
  • Number of moles = 0.467 moles
  • Temperature = 159°C
  • Ideal gas constant, R = 0.08206 L·atm/mol·K

<u>Conversion:</u>

We would convert the value of the temperature in Celsius to Kelvin.

T = 273 + °C

T = 273 + 159

T = 432 Kelvin

To find the pressure of the gas, we would use the ideal gas law;

PV = nRT

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of substance.
  • R is the ideal gas constant.
  • T is the temperature.

Making P the subject of formula, we have;

P = \frac {nRT}{V}

Substituting into the formula, we have;

P = \frac {0.467*0.08206*432}{0.245}

P = \frac {16.5551}{0.245}

<em>Pressure, P = 67.57 atm</em>

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Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
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Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

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