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san4es73 [151]
3 years ago
10

If a solution initially contains 0.260 M HC2H3O2, what is the equilibrium concentration of H3O+ at 25 ∘C? Express your answer in

moles per liter to two significant figures. View Available Hint(s)
Chemistry
1 answer:
arlik [135]3 years ago
8 0

Answer:

2.16 × 10⁻³

Explanation:

Step 1: Given data

Concentration of the acid (Ca): 0.260 M

Acid dissociation constant (Ka): 1.80 × 10⁻⁵

Step 2: Write the acid dissociation equation

HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

Step 3: Calculate the concentration of H₃O⁺ at equilibrium

We will use the following expression.

[H_3O^{+} ]= \sqrt{Ka \times Ca } = \sqrt{1.80 \times 10^{-5} \times 0.260 } = 2.16 \times 10^{-3}

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KATRIN_1 [288]

Answer:

Explanation:

a) Mg + NiCl_2 \rightarrow MgCl_2 + Ni

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and NiCl_2 acts as oxidizing agent.

b) PCl_3 + Cl_2 \rightarrow PCl_5

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

PCl_3 acts as reducing agent and Cl_2 acts as oxidizing agent.

c) C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

C_2H_4 acts as reducing agent and O_2 acts as oxidizing agent.

d) Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and H_2SO_4 acts as oxidizing agent

e) K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

K_2S_2O_3 acts as reducing agent and I_2 acts as oxidizing agent.

f) 3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and HNO_3 acts as oxidizing agent

8 0
3 years ago
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Alexxandr [17]

Answer:

4.21

Explanation:

use Avogadro's number

6.023 x 10^23

multiply this by 7 because you want to find 7 moles :

6.023 x 10^23 x 7 = 4.21

3 0
3 years ago
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Bingel [31]

Answer:

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Explanation:

for pH values we use scientific notation:

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ie:

10^-2 is sol A 10^-5 is sol B

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there's a diff of 1,000 between the solutions.

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Answer:

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