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Neko [114]
2 years ago
14

How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?

Chemistry
1 answer:
Molodets [167]2 years ago
4 0

Answer:

1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm

Explanation:

An ideal gas is a set of atoms or molecules that move freely without interactions. The pressure exerted by the gas is due to the collisions of the molecules with the walls of the container. The ideal gas behavior is at low pressures, that is, at the limit of zero density. At high pressures the molecules interact and intermolecular forces cause the gas to deviate from ideality.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

  • P= 1 atm
  • V= 22.4 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T=273 K

Reemplacing:

1 atm* 22.4 L= n* 0.082 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1 atm* 22.4 L}{0.082 \frac{atm*L}{mol*K} *273 K}

n= 1 mol

Another way to get the same result is by taking the STP conditions into account.

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C (or 273 K) are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<u><em>1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm</em></u>

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Find the weight of HNO_3 present in 20ml, 0.30 N
yanalaym [24]

Answer:

mass of HNO₃ = 0.378 g

Explanation:

Normality = Molarity * number of equivalents

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Read 2 more answers
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Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

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