Asked by Pam
Please tell me which Gas Law equation to use to figure out this problem...
A helium balloon at 25°C, 1 atm, has a volume of 2.88 L. Additional helium is pumped into the balloon. It now contains 1.20 g of He and its new volume is 7.20 L. How many moles of helium gas did the balloon contain originally?
A helium balloon at 25°C, 1 atm, has a volume of 2.88 L. Additional helium is pumped into the balloon. It now contains 1.20 g of He and its new volume is 7.20 L. How many moles of helium gas did the balloon contain originally?
Answers
Answered by
Devron
Let x moles of He=moles of He at 2.88L
1.20g *(1 moles/4.0026g)= moles of He at 7.20L
moles of He/7.20L= x moles of He/2.88L
Solve for x moles of He,
x moles of He=(moles of He/7.20L)*2.88L
1.20g *(1 moles/4.0026g)= moles of He at 7.20L
moles of He/7.20L= x moles of He/2.88L
Solve for x moles of He,
x moles of He=(moles of He/7.20L)*2.88L
Answered by
Pam
Thank you! So the Gas Law I would use would be:
Avogadro’s Law Vi/ni = Vf/nf
I would first need to convert the grams to moles and then pop in the known values and solve for the missing value.
I appreciate your help! I did the calculations based on Avogadro's Law and got the correct answer on my practice test.
Avogadro’s Law Vi/ni = Vf/nf
I would first need to convert the grams to moles and then pop in the known values and solve for the missing value.
I appreciate your help! I did the calculations based on Avogadro's Law and got the correct answer on my practice test.
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