what is the volume occupied by 0.118 mol of helium gas at a pressure of 1.05 atm and a temperature of 322 K?

1 answer:

6 0

The volume occupied by 0.118mol of He gas at a pressure of 1.05 atm and temperature of 322k is calculated by use of ideal gas equation

that is PV = nRT
P(Pressure)1.05 atm
v(volume)=?
n (number of moles)=0.118 mol
T(temperature)= 322K

from equation above volume(v)= nRT/p

volume(v) = (0.118 mol x 0.0821 L.atm/k.mol x322k) / 1.05 atm = 2.97 L

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If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol

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Answer:

$\Delta _RH=4x10^{-3}\frac{kJ}{mol}$

Explanation:

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In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

$\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}$

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

$\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}$