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jeka57 [31]
3 years ago
11

For c2h3o2−, write an equation that shows how the anion acts as a base.

Chemistry
2 answers:
velikii [3]3 years ago
4 0
The synthetic compound Zn(C2H3O2)2 is zinc acetic acid derivation. The molecular composition demonstrates that the mixes incorporate zinc, spoke to by Zn; carbon, spoke to by C; hydrogen, spoke to by H; and oxygen, spoke to by O. Below is the <span>an equation that shows how the anion acts as a base:

</span>

C2H3O2−(aq)+H2O(l)⇌<span>HC2H3O2(aq)+OH−(aq)</span>

Ray Of Light [21]3 years ago
3 0

Explanations:- Acetate ion is a conjugate base of a weak acid(acetic acid). From Bronsted-Lowry concept of acid and base, "Acids are proton donor where as bases are proton acceptors.

To work as a base, acetate ion would be accepting a proton. Water is an amphoteric compound means it could act as an acid as well as a base. So, let's write the reaction of acetate ion with water to show how it acts as a base.

C_2H_3O_2^-(aq)+H_2O(l)\rightleftharpoons HC_2H_3O_2(aq)+OH^-(aq)

In the above reaction, acetate ion has accepted a proton from water means its a base and since water has donated a proton so its an acid.

Acetic acid is the conjugate acid of acetate ion and hydroxide ion is the conjugate base of water.

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96 m/s.

Explanation:

Distance equals speed x time,

12 x 8, so the distance is 96 m/s.

5 0
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What happens when hydrogen reacts with nitrogen at necessary conditions.​
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Why do group 1 cations form precipitates when mixed with hcl?
Xelga [282]
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8 0
3 years ago
Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which ar
Marrrta [24]

Answer:

The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

Explanation:

Let the charge on the cation be q₁

Also let the charge on the anion be q₂

A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C

An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C

The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m

Energy of attraction = \frac{Kq_1q_2}{r^2}

Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂

Energy of attraction = \frac{8.99 X 10^9 (1.602X10^{-19})(4.806 X10^{-19})}{(7.5X10^{-9})^2}

Energy of attraction = 1.231 X 10⁻¹¹ J

Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

8 0
3 years ago
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