A 2.4 L balloon holds 3.7 mol He. If 1.6 mol He are added to the balloon, what is the new volume?
1 answer:
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
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<u>Answer:</u> The for the reaction is -1052.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1)
(2)
The expression for enthalpy of the reaction follows:
Putting values in above equation, we get:
Hence, the for the reaction is -1052.8 kJ.