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Rama09 [41]
4 years ago
11

What is the equilibrium constant of pure water at 25°C?

Chemistry
2 answers:
RSB [31]4 years ago
8 0

Answer:

10↑-14

Explanation:

STatiana [176]4 years ago
5 0

Answer: 1.0 × 10-14

Explanation:

Pure water, represented as

H2O --> [H+] + [OH -]

undergoes a reversible reaction in which both H+ and OH- are generated.

The equilibrium constant for this reaction, called the water dissociation constant, Kw, is 1.0 × 10-14 at 25 °C.

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Answer:

I think its `b

Explanation:

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3 years ago
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These diagrams show two atoms of fluorine and an atom of magnesium.
Diano4ka-milaya [45]

The magnesium atom looses two electrons which are accepted by each of the fluorine atoms. (Option D).

<h3>What is an ionic bond?</h3>

An ionic bond is formed when two ions are formed by the loss or gain electrons. In this case, magnesium has to loose two electrons while fluorine will gain two electrons.

Thus, the correct steps in the formation of an ionic bond between these atoms are A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion (Option D).

Learn more about ionic bond:brainly.com/question/11527546

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6 0
2 years ago
A "shielding gas" mixture of argon and carbon dioxide is sometimes used in welding to improve the strength of the weld. if a gas
kodGreya [7K]
Dalton's Law of Partial Pressures, commonly applied to ideal gases, explains that the partial pressures of individual, non-reacting gases are equal to the total pressure exerted by the gas mixture. The given gas mixture composed of 90% argon and 10% carbon dioxide has the following partial pressures: 3.6 atm for argon and 0.4 atm for carbon dioxide (answer).
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3 years ago
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Sodium is an explosive metal .chlorine is a poisonous gas.you Would expect a compound made of sodium and chlorine to be
inessss [21]
Sodium Chloride, also known as table salt

3 0
3 years ago
Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −
irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a2

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}

The chemical equation for the second dissociation reaction follows:

H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}

The chemical equation for the third dissociation reaction follows:

HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}

To form a buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a of second dissociation process

To calculate the pK_a, we use the equation:

pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})

pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})

We are given:

pK_a2 = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

[HPO_4^{2-}] = concentration of conjugate base

[H_2PO_4^{-}] = concentration of weak acid

Hence, to calculate the pH of the buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a2

3 0
4 years ago
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