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anzhelika [568]
2 years ago
13

Please help I really need it ill give brainliest

Chemistry
1 answer:
madam [21]2 years ago
8 0

Answer:

Sorry, I'm trying to think but i cant explain..

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Need help, please. :)
wolverine [178]

A change of one unit on the pH scale represents a change in the concentration of hydrogen ions by a factor of 10, a change in two units represents a change in the concentration of hydrogen ions by a factor of 100. Thus, small changes in pH represent large changes in the concentrations of hydrogen ions.

6 0
2 years ago
Photochemical smog results from the interaction of pollutants in the presence of
atroni [7]

Water vapor in the air

5 0
2 years ago
Calculate the concentration of hi when the equilibrium constant is 1x10^5
exis [7]
Answer is: concentration of hydrogen iodide is 6 M.

Balanced chemical reaction: H₂(g) + I₂(g) ⇄ 2HI(g).
[H₂] = 0.04 M; equilibrium concentration of hydrogen.
[I₂] = 0.009 M; equilibrium concentration of iodine.
Keq = 1·10⁵.
Keq = [HI]² / [H₂]·[I₂].
[HI]² = [H₂]·[I₂]·Keq.
[HI]² = 0.04 M · 0.009 M · 1·10⁵.
[HI]² = 36 M².
[HI] = √36 M².
[HI] = 6 M.
5 0
3 years ago
The atom with 69 neutrons and 50 protons has a mass number of
GalinKa [24]

Name: Tin

Symbol: Sn

Atomic Number: 50

Atomic Mass: 118.71 amu

8 0
3 years ago
A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil
Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

Now put all the given values in this expression, we get:

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
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