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2 years ago

Many fundamental concepts are at the heart of chemistry. Match the description to the name.

Chemistry

1 answer:

Lisa [10]2 years ago

4 0

Answer:

you dony have a picture for me to match them

Explanation:

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You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H

KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+ pKa1 = 2.14

H2PO4 = HPO4 + H+ pKa2 = 6.86

HPO4 = PO4 + H+ pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

Download odt

7 0

3 years ago

What ion depolarizes the membrane when it diffuses into the axon of a neuron?

Agata [3.3K]

Answer:Sodium ions

Explanation:

The membrane potential of most cells is negative, when sodium ions enters the cells, it can reverse the polarity and makes it positively charge.

7 0

3 years ago

What things can water do while in the gaseous phase?

alisha [4.7K]

Move anywhere and can be anywhere while it is in that phase

8 0

3 years ago

Calculate RMM given Aluminium sulphate given RAM: A1-27; S-32; 0-16

kodGreya [7K]

Answer:

IMISS THE RAGE⁉️⁉️⁉️⁉️⁉️⁉️IMISS THE RAGE⁉️⁉️⁉️⁉️⁉️⁉️IMISS THE RAGE⁉️⁉️⁉️⁉️⁉️⁉️

3 0

3 years ago

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.48 L against a constant pressure

Dahasolnce [82]

Given:
ΔV = 5 - 1.48 L = 3.52 L = 0.00352 m³, th chane in volume
p = 0.818 atm = 0.819*101325 Pa = 82985.2 Pa

By definition, the work done is
$W = \int pdV = (82985.2 \, \frac{N}{m^{2}} )*(0.00352 \, m^{3}) = 292.1 \, J$

Answer: 292.1 J

5 0

3 years ago