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romanna [79]
3 years ago
6

How many objects are in a mole of objects?

Chemistry
1 answer:
lakkis [162]3 years ago
6 0
Your Question: {How many objects are in a mole?}

Helpful Knowledge: (We Know the amount in an object: 12g or C^12)

{A number of objects that are in a mole of objects?}

Well for the question it is pretty easy to answer because a number of objects in One mole would equal 6.02 × 10²³ 

Which 6.02 × 10²³ is an Avogadro's Number. 

So it depends on how many objects you have.

So for every object you have, One mole would equal 6.02 × 10²³. Or 62,000,000,000,000,0000,000,000. Big Number am I right. So that's why we just use 6.02 × 10²³.

Anywho, your answer would be 6.02 x 10²³ x n. 
N would equal the number of objects you're calculating. 

Final Answer: 6.02 x 10²³ x (n) = (Your Answer)

Hope this helps! Have a great day. If you need anything else, feel free to hope right in my inbox. Or comment below. ↓

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An element with a mass number of 11 and an atomic number of 5 has how many<br> neutrons?
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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
3 years ago
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