karlachavez7602 karlachavez7602
  • 16-12-2020
  • Chemistry
contestada

A 22.6 gram sample of krypton gas has a volume of 904 milliliters at a pressure of 2.71 atm. The temperature of the Kr gas sample is:_____

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maacastrobr
maacastrobr maacastrobr
  • 20-12-2020

Answer:

110.78 K

Explanation:

First we convert 22.6 g of krypton gas (Kr) to moles, using its molar mass:

  • 22.6 g Kr ÷ 83.798 g/mol = 0.270 mol Kr

Then we use the PV=nRT formula, where:

  • P = 2.71 atm
  • V = 904 mL ⇒ 904/1000 = 0.904 L
  • n = 0.270 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = ?

2.71 atm * 0.904 L = 0.270 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * T

And solve for T:

  • T = 110.78 K
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