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KatRina [158]
2 years ago
12

calculate the volume occupied by 2.59 mol of helium gas at a pressure of 0.826 atm and a temperature of 356 k. please choose the

correct answer from the following choices, and then select the submit answer button. answer choices
Chemistry
1 answer:
krek1111 [17]2 years ago
8 0

The volume occupied by the given mol of helium gas at the given temperature and pressure is 91.6L.

The Ideal gas Law, also called the overall fuel equation, is the equation of kingdom of a hypothetical ideal gas. It is a superb approximation of the conduct of many gases below many situations, even though it has several limitations.

The precise gas regulation states that PV = NRT, wherein P is absolutely the pressure of a gasoline,volume V is the quantity it occupies, N is the quantity of atoms and molecules inside the fuel, and T is its absolute temperature.

Given;

Amount of helium n = 2.59 mole

Pressure p = 0.826 atm

Temperature T = 356 K

The volume of Helium = ?

PV = nRT

V = nRT/P

V = 2.59 mole *  0.826 atm * 356K/0.826

V = 75.6626/0.826

V = 91.6 L

Learn more about volume here:-brainly.com/question/463363

#SPJ4

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For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea
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Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

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