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3 years ago

Given the ion C2O4-2, what species would you expect to form with each of the following ions?

Chemistry

1 answer:

Ksivusya [100]3 years ago

6 0

Answer:

A. K₂C₂O₄ Potassium oxalate

B. CuC₂O₄ Copper oxalate

C. Bi₂(C₂O₄)₃ Bismuth (III) oxalate

D. Pb(C₂O₄)₂ Lead (IV) oxalate

E. (NH₄)₂C₂O₄ Ammonium oxalate

F. HC₂O₄⁻ Acid oxalate

Explanation:

C₂O₄⁻² → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄ Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O

B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓

This is a precipitate.

C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓

This is a precipitate.

D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓

This is a precipitate.

E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻

HC₂O₄⁻ + H₂O ⇄ C₂O₄⁻² + H₃O⁺ Ka

HC₂O₄⁻ + H₂O ⇄ H₂C₂O₄ + OH⁻ Kb

HC₂O₄⁻ is an amphoteric compound

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2 years ago

Read 2 more answers

(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$