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Sholpan [36]
3 years ago
10

A student has a sample of 1.42 moles of fluorine gas that is contained in a 28.1 L container at 305 K. What is the pressure of t

he sample?
Chemistry
1 answer:
Sphinxa [80]3 years ago
4 0

Answer:

P = 1.26 atm

Explanation:

Given that,

The number of moles, n = 1.42 mol

Volume, V = 28.1 L

Temperature, T = 305 K

We need to find the pressure of the gas. The ideal gas law is as follows :

PV=nRT\\\\P=\dfrac{nRT}{V}

R = 0.0821 L-atm/mol-K

Put all the values,

P=\dfrac{1.42 \times 0.0821\times 305 }{28.1 }\\\\P=1.26\ atm

So, the pressure of the gas is 1.26 atm.

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Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and
andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454&#10;}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0
3 years ago
How much oxygen is typically required to complete a combustion reaction
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Mass of oxygen required = 19.5 moles × 16 g/mol = 312 g.

Explanation:

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What is the percent by mass of water in Na S04.10H20?
dalvyx [7]

Answer:

                    Percent by mass of water is 56%

Explanation:

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Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g

Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g

Mass of Oxygen = O × 14 = 16 × 14 = 224 g

Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g

Mass of Na₂S0₄.10H₂O = 322.24 g

Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...

Mass of water = 10 × 18.02

Mass of water = 180.2 g

Now, we will apply following formula to find percent of water in hydrated compound,

           %H₂O  =  Mass of H₂O / Mass of Hydrated Compound × 100

Putting values,

                                      %H₂O  = 180.2 g / 322.24 g × 100

                                           %H₂O =  55.92 % ≈ 56%

3 0
3 years ago
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