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17 days ago

When the volume of gas is changed from 3.75 to 6.52 L , the temperature will change from 100.0 K to how many k?

Chemistry

1 answer:

eduard [2.4K]17 days ago

3 0

Greetings!

We are dealing with an isobaric transformation, which involves a mass under pressure that retains a consistent pressure; as the temperature rises, the volume enlarges, and conversely, as the temperature lowers, the volume contracts and vice versa.

Here are the pertinent values:

V1 (initial volume) = 3.75 L

V2 (final volume) = 6.52 L

T1 (initial temperature) = 100 K

T2 (final temperature) =? (in Kelvin)

We'll apply the given values to the isobaric transformation formula (Gay-Lussac), check it out:

$\dfrac{V_1}{T_1} =\dfrac{V_2}{T_2}$

$\dfrac{3.75}{100} =\dfrac{6.52}{T_2}$

$3.75*T_2 = 100*6.52$

$3.75\:T_2 = 652$

$T_2 = \dfrac{652}{3.75}$

$\boxed{\boxed{T_2 \approx 173.86\:K}}\Longleftarrow(final\:temperature)\end{array}}\qquad\checkmark$

I hope this is useful, regards... DexteR! =)

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A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition

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Explanation:

Given information:

pH of buffer = 3.75

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Employing the Henderson-Hasselbalch equation for formate buffer:

$pH=pK_a+\log(\frac{[HCOO-]}{[HCOOH]})$

Substituting values into the above equation yields:

$3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1$

$[HCOO-]=[HCOOH]$

Additional given information:

Concentration of formate buffer = 0.100 M

$[HCOO-]+[HCOOH]=0.1$

$[HCOO-]=[HCOOH]=0.05M$

As the buffer volume remains constant, the concentration reflects the moles of formate ions and formic acid.

To find the number of moles based on the specified molarity, utilize the following equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Incorporating these values into the equation gives:

$1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol$

The reaction between formic acid and KOH can be represented as:

$HCOOH+KOH\rightarrow HCOO^-+H_2O$

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$pH=pK_a+\log(\frac{[salt]}{[acid]})$

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$pK_a$ = the negative logarithm of formic acid's dissociation constant = 3.75

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