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ryzh [129]
2 years ago
10

Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction

of aqueous 0.13 M lead (II) nitrate, with 0.19 M potassium carbonate. You may need to consult Appendix E to determine the states of each reactant and product. Assume any insoluble products are completely insoluble.
Chemistry
1 answer:
11Alexandr11 [23.1K]2 years ago
4 0

Answer:

<u>Balanced equation:</u>

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

<u>Ionic equation:</u>

Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)

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Explanation:

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In this experiment, we will be performing a titration with a buret. place the steps in order. 1. record the ph when 0.0 ml of na
77julia77 [94]

I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.

The steps are already in the correct order.

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2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.

3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.

4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.

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Answer:

ΔG° = -533.64 kJ

Explanation:

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Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

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ΔG° = -533.64 kJ

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sergij07 [2.7K]

Answer:

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Explanation:

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