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3 years ago

Give an example of a chemical reaction during which a single substance breaks down. *

Chemistry

1 answer:

RoseWind [281]3 years ago

6 0

Answer:

2NaCl(l) => 2Na(s) + Cl₂(g)

Explanation:

Electrolysis of melt NaCl to form sodium (Na) metal and chlorine (Cl₂) gas.

This is a class of reaction called 'Decomposition Rxns'. For what it's worth, there are 6 distinct compound types that undergo breakdown by decomposition. The acronym C O A C H E can help remember them.

C - Carbonates => CaCO₃(s) => CaO + CO₂

O - Oxides of Metals => 2Fe₂O₃(s) => 4Fe°(s) + 3O₂(g)

A - Acids (Oxoacids) => H₂CO₃(aq) => CO₂(g) + H₂O(l)

C - Chlorates, Bromates & Iodates => 2KClO₃(s) => 2KCl(s) + 3O₂(g)

H - Hydroxides => Ca(OH)₂(s) => CaO(s) + H₂O(l)

E - Electrolysis => see above electrolysis of NaCl(l)

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Explanation:

Given parameters:

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Unknown:

Number of moles of NH₃ = ?

Solution:

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CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.

kupik [55]

Answer:

The correct answer is option a.

Explanation:

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$

Equilibrium concentration cadmium ions = $[Cd^{2+}]=0.0585 M$

Equilibrium concentration fluoride ions = $[F^{-}]=0.117 M$

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$ ..[1]

$NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)$

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0

3 years ago