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IRISSAK [1]
3 years ago
14

What is the ph of a 7.5*10^-3 m H+ solution

Chemistry
2 answers:
Serggg [28]3 years ago
7 0

Answer:

The answer is 2.13

Explanation:

natulia [17]3 years ago
6 0
The pH+ of solution is 3
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Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you
Ludmilka [50]

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

         = 425 mL

The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

7 0
2 years ago
Good chemistry question
NeX [460]
Compared to the charge and mass of a proton an electron has......

A proton has approximately the same mass as..........
5 0
3 years ago
The medium for ocean waves
stepladder [879]
This is a tough one but the answer is water

think of it this way the ocean is water
7 0
2 years ago
Chemical bonds are likely to form when_____
Viefleur [7K]

Answer: When the electrons in an element are more than the required, a bond is formed with other element which has deficiency of electron and in case when electrons are less in numbers the vice versa happens. Such kind of chemical bonds are known as ionic bond.

Explanation:

7 0
2 years ago
Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib
Dvinal [7]

<u>Answer:</u> The value of K_c for the final reaction is 7.16\times 10^{25}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1

<u>Equation 2:</u>  HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2

The net equation follows:

S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=\frac{1}{K_1}\times \frac{1}{K_2}

We are given:

K_1=9.57\times 10^{-8}

K_2=1.46\times 10^{-19}

Putting values in above equation, we get:

K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}

Hence, the value of K_c for the final reaction is 7.16\times 10^{25}

5 0
3 years ago
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