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Mice21 [21]
3 years ago
15

If the pressure on a gas sample is tripled and the absolute temperature is quadrupled, by what factor will the volume of the sam

ple change?
a.12

b.4/3

c.3/4

d.1/3

e.4
Chemistry
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer:

b. 4/3

Explanation:

Given data

  • Initial pressure: P₁
  • Initial volume: V₁
  • Initial temperature: T₁
  • Final pressure: P₂ = 3 P₁
  • Final volume: V₂
  • Final temperature: T₂ = 4 T₁

We can find by what factor will the volume of the sample change using the combined gas law.

\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}}\\\frac{P_{1}.V_{1}}{T_{1}} =\frac{3P_{1}.V_{2}}{4T_{1}}\\V_{1} =\frac{3V_{2}}{4}\\V_{2}=4/3 V_{1}

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Determine the number of representative particles in each of the following:
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(a) 0.294 mol silver = 1.770 * 10^{23}  

(b) 8.98 * 10-3 mol sodium chloride - 5.407 *10^{23}  

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Explanation:

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(b) 8.98 * 10-3 mol sodium chloride - 5.407 *10^{23}  

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3 0
3 years ago
A 500 mL hypertonic saline solution is labeled as consisting of 5.21 % w/w NaCl. Given that the density of salt water is 1.02 g/
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<u>Answer:</u> The molarity of saline solution is 0.909 M

<u>Explanation:</u>

We are given:

5.21 w/w % NaCl

This means that 5.21 grams of NaCl is present in 100 grams of solution

To calculate mass of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.02 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

1.02g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.02g/mL}=98.04mL

To calculate the moalrity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of NaCl = 5.21 g

Molar mass of NaCl = 58.44 g/mol

Volume of solution = 98.04 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{5.21\times 1000}{58.44g/mol\times 98.04}\\\\\text{Molarity of solution}=0.909M

Hence, the molarity of saline solution is 0.909 M

3 0
3 years ago
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