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ivolga24 [154]
2 years ago
5

A sample of neon initially has a volume of 2.80 L at 23 degrees Celsius. What is final temperature, in degrees Celsius, is neede

d to change the volume of the gas to .75 L? Please show your work in order to receive credit.
Chemistry
1 answer:
finlep [7]2 years ago
7 0

The relation between the volume and the temperature of the gas is given by Charles's law. The final temperature of the gas at 0.75 liters is -193.8°C.

<h3>What is Charles's law?</h3>

Charles's law was derived from the ideal gas equation and is used to state the relationship between the temperature and the volume of the gas. With a decrease in volume the temperature decreases.

If the pressure is kept constant then with an increase in temperature the volume of the gas expands. The law is given as,

V₁ ÷ T₁ = V₂ ÷ T₂

Given,

Initial volume (V₁) = 2.80 L

Initial temperature (T₁) = 23 °C = 296.15 K

Final volume (V₂) = 0.75 L

Final temperature = T₂

Substituting the values above as:

T₂ = (V₂ × T₁) ÷ V₁

= 0.75 × 296.15 ÷ 2.80

= 79.325 K

Kelvin is converted as, 79.325K − 273.15 = -193.8°C

Therefore, the final temperature is -193.8°C.

Learn more about Charle's law, here:

brainly.com/question/16927784

#SPJ1

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How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with
Kisachek [45]

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = K_a=6.8\times 10^{-4}

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (6.8\times 10^{-4})

pK_a=4-\log (6.8)

pK_a=3.17

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[NaF]}{[HF]}

Now put all the given values in this expression, we get:

2.60=3.17+\log (\frac{[NaF]}{0.0310})

[NaF]=0.00834M

Now we have to calculate the moles of NaF.

\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole

Now we have to calculate the mass of NaF.

\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g

Therefore, the mass of sodium fluoride added should be 0.105 grams.

4 0
4 years ago
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