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kupik [55]
2 years ago
7

if you needed a buffer that was 2 ph units away from the pka of your assigned buffer, how would you go about making this buffer

Chemistry
1 answer:
labwork [276]2 years ago
8 0

In order to make a buffer that was 2 pH away from the pka, what you have to do is to dilute the buffer.

<h3>How to dilute the buffer.</h3>

If the buffer was made using a base and an acid, then there can be an adjustment through the use of hydrochloric acid.

It can also be adjusted to the value that you want through the use of base such as sodium hydroxide

Read more on buffer here:brainly.com/question/22390063

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If the volume of a gas at 0.5 atm changes from 150 mL to 75 mL, what is the new pressure?
prisoha [69]

Answer:

<h2>1 atm</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we're finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

We have

P_2 =  \frac{0.5 \times 150}{75}  =  \frac{75}{75}  = 1 \\

We have the final answer as

<h3>1 atm</h3>

Hope this helps you

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3 years ago
Is sodium benzoate soluble in NaOH, NaHCO3
krek1111 [17]
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
6 0
3 years ago
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How is 0.00036 written in scientific notation?
natta225 [31]
3.6 x 10^-4
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4 0
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What are the three types of nuclear radiation
Nina [5.8K]
Alpha,Beta and the Gamma
8 0
3 years ago
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Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
Setler [38]

The order of frequency is d < a < c < b

E_{n} = 13.6 * z^{2} / n^{2} eV

where z = atomic mass number

n = energy level

For hydrogen z = 1

Therefore, Energy for n = 1

E_{n} = 13.6 * 1^{2} / 1^{2} eV

     = -13.6 eV

for n = 2

E_{n} = 13.6 * 1^{2} / 2^{2} eV

     = -3.40 eV

for n = 3

 E_{n} = 13.6 * 1^{2} / 3^{2} eV

      = -1.51 eV

for n = 4

 E_{n} = 13.6 * 1^{2} / 4^{2} eV

      = -0.85 eV

for n = 5

 E_{n} = 13.6 * 1^{2} / 5^{2} eV

      = -0.544 eV

n = 2 to n = 4 (absorption)

ΔE = E4 - E2   = -0.85 - (-3.40) = 2.55 eV

n = 2 to n = 1 (emission)

ΔE =  E1 - E2  = -13.6 - (-3.40) = -10.2eV

The negative sign indicates that emission will take place.

n = 2 to n = 5 (absorption)

ΔE = E5 - E2 = -0.544 - (-3.40) = 2.856 eV

n = 4 to n = 3 (emission)

ΔE = E3 - E4 = -1.51 - (-0.85) = -0.66 eV

We know that

E = h * υ

Therefore, Energy is proportional to frequency.

So increasing the order of energy is

E4  < E1  < E3  <  E2

order of frequency is

d < a < c < b

For more information click on the link below:

brainly.com/question/17058029

# SPJ4

8 0
1 year ago
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