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lbvjy [14]
3 years ago
5

What is the morality of 2.50 L of solution that contains 1.85 mol of anhydrous sodium tetraborate?

Chemistry
1 answer:
Lyrx [107]3 years ago
6 0
That would just be 1.85 mol divided by 2.5L. So 0.74 mol/L or 0.74 M.
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Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.30 L and the temperature is increased to 4
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P₂ = 1.12 atm

Explanation:

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\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the new pressure, volume, and temperature. Before plugging the values into the equation, you need to

(1) convert the pressure from mmHg to atm (760 mmHg = 1 atm)

(2) convert the temperatures from Celsius to Kelvin (°C + 273)

The final answer should have 3 sig figs like the given values.

P₁ = 365 mmHg / 760 = 0.480 atm           P₂ = ? atm

V₁ = 5.00 L                                                   V₂ = 2.30 L

T₁ = 20°C + 273 = 293 K                             T₂ = 40°C + 273 = 313 K

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}                                              <----- Combined Gas Law

\frac{(0.480 atm)(5.00 L)}{293 K}=\frac{P_2(2.30 L)}{313 K}                       <----- Insert values

0.00819=\frac{P_2(2.30 L)}{313 K}                                     <----- Simplify left side

2.56 = P_2(2.30L)                                      <----- Multiply both sides by 313

1.12 = P_2                                                  <----- Divide both sides by 2.30

6 0
1 year ago
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