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stepladder [879]
3 years ago
10

How many atoms are in 13.97 liters of water vapor at STP

Chemistry
1 answer:
Arisa [49]3 years ago
5 0
<span>Let's </span>assume that water vapor has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
                                          n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.

According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span>                                                                = 3.758 x 10</span>²³<span>

</span>
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The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(l)ΔH=
castortr0y [4]

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The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

Explanation:

<u>Step 1</u>: Data given

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<u>Step 2:</u> Multiply equations

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Multiply the second equation by 6:

12 NO(g) + 6 O2(g) → 12 NO2(g) ΔH = −678 kJ

Multiply the third equation by 4:

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<u>Step 3:</u> Get the equations together

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12 NH3(g) + 21 O2(g) → 14 H2O(l) + 8 HNO3(aq) + 4 NO(g) ΔH = −3955 kJ

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−3955 kJ/8 moles HNO3= −494 kJ

The total energy change for the production of one mole of aqueous nitric acid is −494 kJ

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Explanation:

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