3. Oxalic acid, C H20, is a toxic substance found in rhubarb leaves. When mixed with sufficient quantities of a strong base, this weak diprotic acid loses two protons to form a polyatomic ion
called oxalate, C2022. Write a balanced equation that describes the reaction between oxalic acid
and sodium hydroxide.

COOHCOOH + 2OH⁻ ⇄  C₂O₄²⁻ + 2H₂O

Explanation:

The reaction of oxalic acid with a strong base like sodium hydroxide is the following:

COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O    (1)

In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:

COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O      (2)  

The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):

COOHCOOH + 2OH⁻ ⇄  C₂O₄²⁻ + 2H₂O              

I hope it helps you!  

the answer is   c.) 1575 j.

explanation:

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be