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kodGreya [7K]
3 years ago
13

when 6.0 mol of oxygen are confined in a 36L vessel at 196°c, the pressure is 8atm. what is the new pressure for oxygen expands

at constant temperature to fill 48L
Chemistry
1 answer:
kykrilka [37]3 years ago
8 0

<u>Answer:</u> The new pressure for oxygen gas is 6 atm.

<u>Explanation:</u>

To calculate the new pressure of the gas, we use the equation given by Boyle's Law.

This law states that the pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

Mathematically,

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are the initial pressure and volume of the gas.

P_2\text{ and }V_2 are the final pressure and volume of the gas.

We are given:

P_1=8atm\\V_1=36L\\P_2=?atm\\V_2=48L

Putting values in above equation, we get:

8atm\times 36L=P_2\times 48L\\\\P_2=6atm

Hence, the new pressure for oxygen gas is 6 atm.

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A 2.5 mol sample of phosphorus pentachloride, PCl5 dissociates at 160C and 1.00atm to give 0.338 mol of phosphorus trichloride a
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Explanation:

Moles of phosphorus pentachloride present initially = 2.5 mol

Moles of phosphorus trichloride at equilibrium = 0.338 mol

PCl_5\rightleftharpoons PCl_3+Cl_2

Initially

2.5 mol      0    0

At equilibrium:

(2.5 - x) mol      x     x

So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol

Mass of 0.338 moles of  phosphorus trichloride at equilibrium:

= 0.338 mol × 137.5 g/mol = 46.475 g

Moles of phosphorus pentachloride present at equilibrium :

= (2.5 - 0.338) mol = 2.162 mol

Mass of 2.162 moles of  phosphorus pentachloride at equilibrium:

= 2.162 mol × 208.5 g/mol = 450.777 g

Moles of chloride gas present at equilibrium : 0.338 mol

Mass of 0.338 moles of chloride gas at equilibrium:

= 0.338 mol × 71 g/mol = 23.998 g

3 0
3 years ago
For the reaction shown here, 3.5 mola is mixed with 5.9 molb and 2.2 molc. what is the limiting reactant?3a+2b+c→2d
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<span>For equation A + 3B + 2C ---> 2D, 1 mole of A will produce 2 moles of D 3 moles of B will produce 2 moles of D, so 1 mole of B will produce 2/3 moles of D 2 moles of C will produce 2 moles of D, so 1 mole of C will produce 1 mole of D If only 1 mole of B is present, only 2/3 moles of D can be produced. This is regardless of the number of moles of A and C. B is the limiting reactant and the maximum number of moles of D expected is 2/3.</span>
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3 years ago
A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif
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Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = [A_o]

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = [A]=0.06[A_o]

Half life of the carbon-14 = t_{1/2}=5730 years

k=\frac{0.693}{t_{1/2}}

[A]=[A_o]\times e^{-kt}

[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}

0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}

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t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0
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