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The number of grams of helium in a balloon at a pressure of 99.8 kPa, a temperature of 301 K, and a volume of 0.785 L would be:

A. 814 g
B. 0.125 g
C. 0.278 g
D. 337 g

Answer :

use formula
pV=nRT
in formula n=m/M
you have to find m



P=99.8/101.325
v=0.785
R=0.0821
T=301

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Rewritten by : Jeany

The number of grams of helium in a balloon at a pressure of 99.8 kPa, a temperature of 301 K, and a volume of 0.785 L would be 0.125g.

How to calculate mass?

The mass of a substance can be calculated by multiplying the number of moles by its molar mass.

However, the number of moles of helium must be calculated as follows:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • n = no of moles
  • R = gas law constant
  • T = temperature

0.985 × 0.785 = n × 0.0821 × 301

0.773 = 24.7n

n = 0.773 ÷ 24.7 = 0.031moles

mass of He = 0.031 × 4 = 0.125g

Therefore, the number of grams of helium in a balloon at a pressure of 99.8 kPa, a temperature of 301 K, and a volume of 0.785 L would be 0.125g.

Learn more about mass at: https://brainly.com/question/19694949

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