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Sidana [21]
4 years ago
5

Taking into account the vapor pressure of water, how many moles of hydrogen gas, n, are present in 345 ml at 752 torr and 29 ∘c?

the value of the gas constant r is 0.08206 l⋅atm/(mol⋅k). you may also find the conversion
Chemistry
1 answer:
SVEN [57.7K]4 years ago
3 0
To make things simple, assume ideal gas. Then, we use the ideal gas equation.

PV = nRT

Since we are finding for n, the equation should be rearranged to

n = PV/RT

Make sure you are consistent with your units. If you use R = <span>0.08206 l⋅atm/(mol⋅k), the pressure should be:
P = 752 torr (1 atm/760 torr) = 0.98947 atm
The volume should be:
V = 345 mL (1 L/1000 mL) = 0.345 L
The temperature should be:
T = 29 + 273 = 302 K

Thus,

n = (</span>0.98947 atm)(0.345 L)/(0.08206 l⋅atm/(mol⋅k))(302 K)
<em>n = 0.01377 moles of hydrogen gas</em>
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The above-balanced equation can be solved algebraically to obtain the required Kgoal value.

Adding given equations 1, 2 and 3 we obtain the required equation.

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K_1 \times K_2 \times K_3 = K_{goal}

K_{goal}= 4.10 \times 10^{-31} \times 7.40 \times 10^{-26} \times 1.06 \times 10^{-10}

K_{goal}= 3.22\times 10 ^{-66}

Part B:

The required equation is balanced, Now

Let.

P₄(s)+6Cl₂(g) ⇌ 4PCl₃(g), K₁=2.00×10¹⁹ ------------------------------------ (a)

PCl₅(g) ⇌ PCl₃(g)+Cl₂(g), K₂=1.13×10⁻² --------------------------------------- (b)

By multiplying equation 2 by 4 and subtracting equation 1 from it, we get

4PCl₅(g) ⇌ P₄(s)+10Cl₂(g)

The Kgoal for the above equation is the product of four times K₂ and inverse K₁ according to the applied operation. Mathematically,

K_{goal}= 4K_2 \times \frac{1}{K_1}

K_{goal}= 4(1.13\times10^{-2}) \times \frac{1}{2.00\times10^{19}}

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Read 2 more answers
Calculate the root mean square velocity of nitrogen molecules at 25°c. a) 729 m/s b) 515 m/s c) 149 m/s d) 297 m/s
Elena L [17]

Answer:

root mean square velocity of nitrogen at 25°c is 515 m/s

correct option is b) 515 m/s

Explanation:

given data

temperature = 25°c = 25 + 273 = 298 K

to find out

root mean square velocity of nitrogen molecules

solution

we know that root mean square velocity of gas is Vrms = \sqrt{\frac{3RT}{M}}      .......................1

mass of gas = M

universal gas constant = R

temperature = T

and we know mass of nitrogen = 28 g = 28 × 10^{-3} kg

Vrms = \sqrt{\frac{3RT}{M}}

Vrms = \sqrt{\frac{3(8.314)298}{28*10^{-3}}}

Vrms = 515.22 m/s

root mean square velocity of nitrogen at 25°c is 515 m/s

correct option is b) 515 m/s

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