What is the mass in grams of 2.21 mole of Ca3(PO4)4?
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Answer:
the final mole of the flexible container = 12.92 moles
Explanation:
Given that :
initial volume of a flexible container = 6.13 L
initial mole of a flexible container = 6.51 mol
final volume of a flexible container = 18.3 L
final mole of a flexible container = ???
Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Therefore,
n = 19.43
19.43 = 6.51 + n₂
n₂ = 19.43 - 6.51
n₂ = 12.92 moles
Thus; the final mole of the flexible container = 12.92 moles