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3. At what temperature in Celsius will 9.8 grams of NH3 will exert a pressure of 98.3 KPa at a volume of 21.0 Liters?

Sagot :

maacastrobr

Answer:

156.78 °C

Explanation:

First we convert 9.8 grams of NH₃ into moles, using its molar mass:

  • 9.8 g ÷ 17 g/mol = 0.576 mol

Then we can use the PV=nRT formula to calculate the temperature, where:

  • P =  98.3 kPa ⇒ 98.3/101 = 0.967 atm
  • V = 21.0 L
  • n = 0.576 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = ?

We input the data:

  • 0.967 atm * 21.0 L = 0.576 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * T

And solve for T:

  • T = 429.94 K

Finally we convert 429.94 K to °C:

  • 429.94 - 273.16 = 156.78 °C
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