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sammy [17]
4 years ago
7

A sample of a gas has a pressure of 248mm Hg when the volume is 1.05L. What is the pressure of the gas when the volume is 3.98L?

Chemistry
2 answers:
ozzi4 years ago
8 0
P1V1 = P2V2
Convert 248mm Hg to atm
248/760 = 0.326 atm
(0.326 atm • 1.05L) / 3.98L = 0.0861 atm
slamgirl [31]4 years ago
6 0

So Boyle's Law equation is going to be used to solve this problem.

Boyle's Law: (P1)(V1)=(P2)(V2)

So, based on the question. Let's first note down what is given.

P1= 248mm Hg

V1= 1.05L

P2= ? (<-- What we are going to be solving for)

V2= 3.98L

Based on what we know, we can just simply plug it into the equation.

So...

1. (248 mmHg) (1.05L) = (P2) (3.98L)

2. (248 mm Hg) (1.05L)

   -----------------------------    =     P2

        (3.98L)

3. P2 = 65.4 mm Hg is your answer

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Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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