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Salsk061 [2.6K]
2 years ago
5

A 4 L sample of gas at 298 K and 2 atm contains 0.250 mol of gas. If we add another 0.250 mol of gas at the same pressure and te

mperature, what is the final total volume of gas?
Chemistry
1 answer:
andrey2020 [161]2 years ago
4 0

Answer:

V_2=8L

Explanation:

Hello there!

In this case, considering the Avogadro's gas law, which treats the volume and moles in a directly proportional way via:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

Which can be solved for the final volume, V2, as shown below:

V_2=\frac{V_1n_2}{n_1}

Thus, by plugging in the given data, we obtain:

V_2=\frac{4L*(0.250mol+0.250mol)}{0.250mol}\\\\V_2=8L

Regards!

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The answer is #2// gas
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Fill in the coefficients that will balance the following reaction:
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Explanation:

Given equation;

      NaC₂H₃O₂   +  Fe₂O₃   →   Fe(C₂H₃O₂)₃  +  Na₂O

To find the coefficient that will balance this we equation, let us set up simple mathematical algebraic expressions that we can readily solve.

Let us have at the back of our mind that, in every chemical reaction, the number of atom is usually conserved.

      aNaC₂H₃O₂   +  bFe₂O₃   →   cFe(C₂H₃O₂)₃  +  dNa₂O

a, b, c and d are the coefficients that  will balance the equation.

 conserving Na;    a  = 2d

                       C:     2a  = 6c

                       H:     3a  = 9c

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let a = 1

    solving:

      2a = 6c

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          c = \frac{2}{6}  = \frac{1}{3}

      2b = c

          b = \frac{1}{2}  x \frac{1}{3} = \frac{1}{6}

     d = 2a + 3b - 6c = 2(1 ) + (3 x\frac{1}{6}  )  - (6  x  \frac{1}{3}) = \frac{1}{2}

Now multiply through by 6

  a = 6, b = 1, c = 2 and d = 3

             

                6NaC₂H₃O₂   +  Fe₂O₃   →   2Fe(C₂H₃O₂)₃  +  3Na₂O

learn more:

Balanced equation brainly.com/question/9325293

#learnwithBrainly

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Answer:

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The answer would be D because from my research it's the only one that didn't have a catalyst 
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