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babunello [35]
3 years ago
13

What mass of nitrogen is needed to fill an 855 L tank at STP?

Chemistry
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

  • P= 1 atm
  • V= 855 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 273.15 K

Replacing:

1 atm* 855 L= n* 0.082 \frac{atm*L}{mol*K} * 273.15 K

Solving:

n=\frac{1 atm* 855 L}{0.082\frac{atm*L}{mol*K} *273.15 K }

n= 38.17 moles

Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

mass=\frac{38.17 moles*28 grams}{1 mole}

mass= 1,068.76 grams

<u><em> It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.</em></u>

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      <u><em>Explanation</em></u>

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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
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Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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The partial pressure of Br_2 = 2.0\times 10^{-4}atm

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The balanced equilibrium reaction is,

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Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

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The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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