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cestrela7 [59]
3 years ago
5

If 9.9L of helium are in a tire at 303k,how many liters will be present at 403k if the pressure is held constant.

Chemistry
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

Final volume, V2 = 13.18 Liters

Explanation:

<u>Given the following data;</u>

Initial volume = 9.9 L

Initial temperature = 303 K

Final temperature = 403 K

To find the final volume, we would use Charles law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles' law is given by the formula;

VT = K

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Where;

  • V1 and V2 represents the initial and final volumes respectively.
  • T1 and T2 represents the initial and final temperatures respectively.

\frac{V1}{T1} = \frac{V2}{T2}

Making V2 as the subject formula, we have;

V_{2}= \frac{V1}{T1} * T_{2}

V_{2}= \frac{9.9}{303} * 403

V_{2}= 0.0327 * 403

<em>Final volume, V2 = 13.18 Liters</em>

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If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?
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The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

                       pH = pKa + log10 ([A–]/[HA])

Here, 100 mL  of  0.10 m TRIS buffer pH 8.3

                 pka = 8.3

         0.005 mol of TRIS.

∴  8.3 = 8.3 + log \frac{[0.005]}{[0.005]}

<em>    </em>inverse log 0 = \frac{[B]}{[A]}

   \frac{[B]}{[A]} = 1

Given; 3.0 ml of 1.0 m hcl.

           pka = 8.3

           0.003 mol of HCL.

pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69

Therefore, the new pH is 7.69.

Learn more about pH here:

brainly.com/question/24595796

#SPJ1

 

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