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GuDViN [60]
2 years ago
10

Calculate the [H+] for a solution with a pOH of 4.75

Chemistry
1 answer:
S_A_V [24]2 years ago
7 0

Answer:

Explanation:

A. 1.3 x 10-10 M

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Zanzabum
Bonds between atoms break and new ones form and energy is either given out or taken in
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2 years ago
All solutions<br>are mixtures; but<br>all mixtures are<br>not solutions.<br>why?​
aleksklad [387]

All solutions are mixtures of two or more substances, but unless the mixture has a homogeneous distribution of solutes in the solvent, then the mixture is not a solution. Therefore, all mixtures are not solutions.

5 0
2 years ago
How much heat is required to raise the temperature of 81.0 g of water from its melting point to its boiling point?
Dovator [93]

Answer:

Specific heat of water = 33.89 KJ

Explanation:

Given:

mass of water = 81 gram

Initial temperature = 0°C

Final temperature = 100°C

Specific heat of water = 4.184

Find:

Required heat Q

Computation:

Q = Mass x Specific heat of water x (Final temperature - Initial temperature)

Q = (81)(4.184)(100-0)

Q = 33,890.4

Specific heat of water = 33.89 KJ

6 0
3 years ago
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

NH3 + H2O(l) ⇄ NH4+ + OH-

Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

pH:

As pOH = -log [OH-]

pOH = 2.48

pH = 14 - pOH

<h3>pH = 11.52</h3>
5 0
2 years ago
Sea water is an electrolyte.<br> True<br> False
antoniya [11.8K]
If you are talking about just pure regular water, the answer is false. BUT, some salts dissolved IN WATER, can act as electrolytes. But regular water, no.
8 0
3 years ago
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