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iVinArrow [24]
3 years ago
14

A 3.00-l flask is filled with gaseous ammonia, nh3. the gas pressure measured at 26.0 ∘c is 1.55 atm . assuming ideal gas behavi

or, how many grams of ammonia are in the flask?
Chemistry
1 answer:
Phantasy [73]3 years ago
3 0

Answer: -

A 3.00-l flask is filled with gaseous ammonia, NH₃. the gas pressure measured at 26.0 ∘c is 1.55 atm . assuming ideal gas behavior, 3.23 grams of ammonia are in the flask

Explanation: -

Volume V = 3.00 L

Pressure P = 1.55 atm

Temperature T = 26.0 °C + 273 = 299 K

We know the value of Universal gas constant R = 0.082 L atm K−1

mol−1

We use the ideal gas equation

PV = nRT

Number of moles of Ammonia n = \frac{PV}{RT}

= \frac{1.55 atm x 3.00 L}{0.082 L atm K−1
 x 299 K}

= 0.19 mol

Molar mass of NH₃ = 14 x 1 + 1 x 3 = 17 g / mol

Mass of NH₃ = Molar mass of NH₃ x number of moles of NH₃

= 17 g / mol x 0.19 mol

= 3.23 g

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an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem
OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

P\cdot V = n\cdot R\cdot T,

\displaystyle V = \frac{n\cdot R\cdot T}{P}.

where

  • P is the pressure of the gas,
  • V is the volume of the gas,
  • n is the number of moles of particles in this gas,
  • R is the ideal gas constant, and
  • T is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of R will be 8.314 if P, V, and T are in SI units. Convert these values to SI units:

  • P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa;
  • V shall be in cubic meters, \rm m^{3};
  • T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K.

Apply the ideal gas law:

\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}.

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3 years ago
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