What would be the mass of a 40 cm3 piece of manganese?
2 answers:
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For this question, lets apply Avagadro's law
when Pressure and temperature are constant, the volume occupied is directly proportional to the number of moles of gases.
where V-volume, n-number of moles and k - constant
Therefore at 2 instances
where V1 and n1 are for 1st instance
and V2 and n2 are for 2nd instance
therefore
V1 = 2.4 L
n1 = 3.7 mol
n2 = 3.7 + 1.6 = 5.3 mol
since more He moles are added at the 2nd instance its the sum of the moles.
V2 needs to be calculated
V2 = 2.4 x 5.3 / 3.7
= 3.4 L
Answer is 1st option 3.4 L
when Pressure and temperature are constant, the volume occupied is directly proportional to the number of moles of gases.
where V-volume, n-number of moles and k - constant
Therefore at 2 instances
where V1 and n1 are for 1st instance
and V2 and n2 are for 2nd instance
therefore
V1 = 2.4 L
n1 = 3.7 mol
n2 = 3.7 + 1.6 = 5.3 mol
since more He moles are added at the 2nd instance its the sum of the moles.
V2 needs to be calculated
V2 = 2.4 x 5.3 / 3.7
= 3.4 L
Answer is 1st option 3.4 L