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jarptica [38.1K]
3 years ago
13

5.0 L vessel holds 3.0 mol n2, 2.0 mol f2, and 1.0 mol h2 at 273 K. What is the partial pressure of fluorine?

Chemistry
1 answer:
xz_007 [3.2K]3 years ago
4 0

Answer:- 8.97 atm.

Solution:- Moles of each gas, volume of the vessel and temperature are given. So, we could easily use the partial pressure of each gas using ideal gas law equation.

PV = nRT

P is the pressure, V is the volume, n is the moles of the gas, R is universal gas constant and T is kelvin temperature.

The equation is rearranged for the pressure as:

P=\frac{nRT}{V}

To calculate the partial pressure of fluorine we will use its moles.

n = 2.0 mol

V = 5.0 L

T = 273 K

R = 0.0821\frac{atm.L}{mol.K}

P = ?

Let's plug in the values and calculate the partial pressure of fluorine.

P=\frac{2.0mol*0.0821\frac{atm.L}{mol.K}*273K}{5.0L}

P = 8.97 atm

So, the partial pressure of fluorine gas is 8.97 atm.

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Hi May I know how to balance this
almond37 [142]

Answer:

  2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

Explanation:

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For Ba: 3a = d

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Expressing everything in terms of b and c, we get ...

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From the second, b = 6c, so we have ...

  a = 2c

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Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

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Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

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Explanation:

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