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erastova [34]
3 years ago
5

7. What is the total pressure of the following mixture of gases in a 20.0 L container at 298 K: 12.0 g He,

Chemistry
1 answer:
zloy xaker [14]3 years ago
5 0

Answer:

P(total) = 14.81 atm

Explanation:

According to the Dalton law of partial pressure,

The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.

P(total) = P1 + P2 + P3+ .....+ Pn

Given data:

Volume of container = 20.0 L

Temperature = 298 K

Mass of He = 12.0 g

Moles of hydrogen = 4.00 mol

Pressure of Ne = 6.25 atm

Solution:

Moles of helium:

Number of moles = mass/molar mass

Number of moles = 12 g/ 4 g/mol

Number of moles = 3 mol

Pressure of helium:

PV = nRT

P = nRT/ V

P = 3 mol × 0.0821 atm.L. mol⁻¹.k⁻¹ × 298 K / 20 L

P = 73.4 atm / 20

p = 3.67 atm

Pressure of hydrogen:

PV = nRT

P = nRT/ V

P = 4 mol × 0.0821 atm.L. mol⁻¹.k⁻¹ × 298 K / 20 L

P = 97.86 atm / 20

P = 4.89 atm

Total pressure:

P(total) = P(He) + P(Ne) + P(H₂)

P(total) = 3.67 atm + 6.25 atm + 4.89 atm

P(total) = 14.81 atm

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Answer:

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Explanation:

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Which is a complex molecule?<br><br> a. H2<br><br> b. Ne<br><br> c. Co<br><br> d. H20
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The chemical formula for table sugar is C 12 H 11 O 22 . What can you tell from this formula?
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Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N
Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of NaN_3

\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}

Molar mass of NaN_3 = 65.01 g/mole

\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 2 mole of NaN_3 react to give 3 mole of N_2

So, 0.311 moles of NaN_3 react to give \frac{0.311}{2}\times 3=0.466 moles of N_2

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = ?

n = number of moles N_2 = 0.466 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 22^oC=273+22=295K

Putting values in above equation, we get:

1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K

V=11.3L

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0
3 years ago
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