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Sedbober [7]
3 years ago
12

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2 If you start with 26.0 mL of a 0.904 M solution of NaIO4, and then

dilute it with water to 500.0 mL, what is the concentration of H4IO−6 at equilibrium?
Chemistry
1 answer:
Oksi-84 [34.3K]3 years ago
5 0

Answer:

0.744 M

Explanation:

IO⁻⁴(aq) + 2H₂O(l) ⇌ H₄IO⁻⁶(aq)

Kc = 3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

First let's<u> calculate the new concentration of IO⁻⁴ at equilibrium</u>:

0.904 M * 26.0 mL / 500.0 mL = 0.047 M = [IO⁻⁴]

Now we can<u> calculate [H₄IO⁻⁶] using the formula for Kc</u>:

3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

3.5×10⁻²= [H₄IO⁻⁶] / 0.047 M

[H₄IO⁻⁶] = 0.744 M

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Convert 3 mol of Co, to grams.
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176.7996 grams

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An aqueous sodium acetate, NaC2H3O2 , solution is made by dissolving 0.395 mol NaC2H3O2 in 0.505 kg of water. Calculate the mola
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<u>Answer:</u> The molality of NaC_2H_3O_2 solution is 0.782 m

<u>Explanation:</u>

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:

\text{Molality of solution}=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} .....(1)

Given values:

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Putting values in equation 1, we get:

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Hence, the molality of NaC_2H_3O_2 solution is 0.782 m

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