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vovangra [49]
1 year ago
13

What is the ph of a solution containing 0.01 m acetic acid (pka = 4.7) and 0.1 m sodium acetate?

Chemistry
1 answer:
Pepsi [2]1 year ago
8 0

The ph of a solution is 3.7

Solution:

According to the equaiton of Henderson-Hasselback

      pH= pKa+ log(salt/acid)

here it is given the value of

      pKa= 4.7

So,

      pH = pKa+ log(0.1/0.01)

            = 4.7 + log(0.1)

            = 4.7–1

            = 3.7.

The following problem illustrates how the Henderson-Hasselbalch equation can be used to determine how much acid and conjugate base should be combined to create a buffer solution with a specific pH.

To learn more click the given link

brainly.com/question/13423434

#SPJ4

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How many calories is 52 J of energy?
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Answer:

1J = 0.238846cal

52J = 12.42cal

5 0
3 years ago
The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge fo
RSB [31]

Answer:

- 1.602 x 10⁻¹⁹coulombs

Explanation:

Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.

LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹)  because it is  already a common factor to all.

The LCM  is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.

4 0
4 years ago
What will be the new volume if 125 mL of He gas at 100 degree Celsius and .979 atm is cooled to 26 degree Celsius and the pressu
ycow [4]
Use Boyle's Law of Pressure: P1 x V1 = P2 x V2. Givens: P1=0.9 atm V1=                4 P2= 0.9 atm Find: V2 Equation: 0.9 atm x 4 x 4 L = 0.20 atm x V2Solve: 36 atmL= 0.20 atm x V2 18 : = V2 Short answer: The new volume is 104 ml. 
6 0
4 years ago
The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma
Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})}  \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}

By replacing the corresponding values we obtain:

\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0
3 years ago
A scientist is unsure about the accuracy of her experiment. She has checked her equipment and found it to be in good working ord
polet [3.4K]

Answer:

D

Explanation:

The correct thing to do in this case would be to <u>repeat the experiment.</u>

The scientist would need to repeat the experiment in order to double-check the accuracy. If the accuracy is indeed doubtful, he/she can be able to trace the source of the error by repeating the experiment.

The correct option is D.

8 0
3 years ago
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