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riadik2000 [5.3K]
3 years ago
13

A 2.0 L flexible container holds 5.0 moles of oxygen (O2) gas. An additional 15.0 moles of nitrogen gas (N2) is added to the con

tainer. What is the new volume of the container?
a. 8.0 L
b. 0.13 L
c. 0.50 L
d. 6.0L
Chemistry
2 answers:
professor190 [17]3 years ago
7 0

The  new  volume of the container is 8.0 L


 calculation


total  volume=    volume of  O2 +  volume of N2

 volume of  O2 =  2.0 l

find the volume of N2  which is calculated as below


if 2.0 L  =  5.0 moles

   ?  L  = 15.0  moles

by cross multiplication

= (15.0 moles  x  2.0 L) / 5.0 mole = 6.0 L


total volume ( new volume ) =  6.0 L +  2.0 L = 8.0 L


frozen [14]3 years ago
4 0
Answer 8.0 L.

2.0L / 5.0 moles = x / 20.0 => x = 20 / 5 * 2 = 8
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Answer:

No.

Explanation:

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3 years ago
A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What
NeTakaya

Answer:

Approximately 1.854\; \rm mol\cdot L^{-1}.

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of \rm Sr, \rm O, and \rm H on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

  • \rm Sr: 87.62.
  • \rm O: 15.999.
  • \rm H: 1.008.

Calculate the formula mass of \rm Sr(OH)_2:

M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}.

<h3>Number of moles of strontium hydroxide in the solution</h3>

M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1} means that each mole of \rm Sr(OH)_2 formula units have a mass of 121.634\; \rm g.

The question states that there are 10.60\; \rm g of \rm Sr(OH)_2 in this solution.

How many moles of \rm Sr(OH)_2 formula units would that be?

\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}.

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There are 8.71467\times 10^{-2}\; \rm mol of \rm Sr(OH)_2 formula units in this 47\; \rm mL solution. Convert the unit of volume to liter:

V = 47\; \rm mL = 0.047\; \rm L.

The molarity of a solution measures its molar concentration. For this solution:

\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}.

(Rounded to four significant figures.)

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Answer:

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Explanation:

                                              NH4SH (s)      <-->            NH3 (g) + H2S (g)

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