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10.0 grams of argon and 20.0 grams of neon are placed in a 1040.0 ml container at 31.5 °c. the partial pressure of neon is ________ atm.

Respuesta :

The partial pressure of neon is 24.0498 atm.

To calculate the partial pressure of neon, we can use the ideal gas equation,

PV = nRT

Pne = nRT / V

Number of moles of neon = Mass of neon / Mass of one mole of neon

n = 20 / 20

n = 1 mole

R = 0.0821 L atm K-1 mol-1

T = 31.5 °C

T = 273.15 + 31.5

  = 304.65 K

V = 1040.0ml

Pne = 1 * 0.0821 * 304.65 / 1040.0 * 10-3

Pne = 24.0498 atm

The partial pressure of neon is 24.0498 atm.

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