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erica [24]
3 years ago
13

Is there such thing as a pure mixture?

Chemistry
2 answers:
irina1246 [14]3 years ago
6 0
Matter can be broken down into two categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.
prisoha [69]3 years ago
3 0
Pure substances are further broken down into elements and compounds. Mixture are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.
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Impressionist paintings are a non-example of electric circuits.

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What is scientific literacy?<br><br> Why is it important?
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WILL GIVE BRAINLIEST IF YOU EXPLAIN (this is 7th grade Science / Chemistry btw):
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2 chlorine atoms.

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Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha
OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

KH_2PO_4 acts as a weak acid and Na_2HPO_4 as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

7.00=6.86+log(\frac{base}{0.10})

7.00-6.86=log(\frac{base}{0.10})

0.14=log(\frac{base}{0.10})

Taking antilog:

10^0^.^1^4=\frac{base}{0.10}

1.38=\frac{base}{0.10}

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is Na_2HPO_4 required to make the buffer is 0.138M.

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