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Snowcat [4.5K]
3 years ago
15

What is the pH of solution if the concentration of [H3O+] = 0.000558?

Chemistry
1 answer:
lina2011 [118]3 years ago
7 0

Answer:

3.25

Explanation:

pH = -log[H3O+]

-log(0.000558)=3.25

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Write the chemical equations involved in this experiment and how that the rate of disappearance of [S2O8^2-] is proportional to
Ne4ueva [31]

S₂O₈²⁻

(aq) + 2I⁻

(aq) → I₂(aq) + 2SO₄

²⁻(aq)

2S₂O₃²⁻

(aq) + I₂(aq) → S₄O₆²⁻

(aq) + 2I⁻

(aq)

<u>Explanation:</u>

S₂O₈²⁻

(aq) + 2I⁻

(aq) → I₂(aq) + 2SO₄

²⁻(aq)

To measure the rate of this reaction we must measure the rate of concentration change of one of  the reactants or products. To do this, we will include (to the reacting S₂O₈

²⁻  and I⁻

i) a small amount of sodium thiosulfate, Na₂S₂O₃,

ii) some starch indicator.

The added Na₂S₂O₃ does not interfere with the rate of above reaction, but it does consume the I₂  as soon as it is formed.

2S₂O₃²⁻

(aq) + I₂(aq) → S₄O₆²⁻

(aq) + 2I⁻

(aq)

This reaction is much faster than the previous, so the conversion of I2 back to I⁻  is  essentially instantaneous.

rate = \frac{dI2}{dt} = \frac{1/2 [S2O3^2^-]}{t}

5 0
3 years ago
HELP ME PLEASE ASAP. I'LL MARK YOU BRAINLIEST <br> ITS NUMBER 23.
Ivanshal [37]

Answer:

The answer is A

Explanation:

5 0
3 years ago
Define satt and give an example?​
Veseljchak [2.6K]

Answer:

<h2><em>Heyy</em><em>.</em><em>.</em><em>Here</em><em> </em><em>is</em><em> </em><em>your</em><em> </em><em>answer</em><em>.</em><em>.</em></h2>

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7 0
2 years ago
How many electrons does nitrogen (N) need to gain to have a stable outer electron shell
earnstyle [38]
It need "3 electrons" to have a stable electronic configuration. 

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Hope this helps!
7 0
3 years ago
Read 2 more answers
Calculate the molar solubility of CaF2 in a 0.25 m solution of NaF(aq).
Kipish [7]

Answer:

6.4 × 10^-10 M

Explanation:

The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).

CaF2 will dissociate as follows:

CaF2 ⇌Ca2+ + 2F-

1 mole of Calcium ion (x)

2 moles of fluorine ion (2x)

NaF will also dissociate as follows:

NaF ⇌ Na+ + F-

Where Na+ = 0.25M

F- = 0.25M

The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M

Ksp = {Ca2+}{F-}^2

Ksp = {x}{0.25}^2

4.0 × 10^-11 = 0.25^2 × x

4.0 × 10^-11 = 0.0625x

x = 4.0 × 10^-11 ÷ 6.25 × 10^-2

x = 4/6.25 × 10^ (-11+2)

x = 0.64 × 10^-9

x = 6.4 × 10^-10

Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M

8 0
3 years ago
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