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4 years ago

5

Sodium chloride reacts with copper sulfate to produce sodium sulfate and copper chloride. 2 upper N a upper C l (a q) plus upper

C u Uppe S upper O subscript 4 (a q) right arrow upper N a subscript 2 upper S upper O subscript 4 (a q) plus upper C u upper C l subscript 2 (s). This equation represents a synthesis reaction. decomposition reaction. single replacement reaction. double replacement reaction.

2 answers:

V125BC [204]4 years ago

8 0

Aqueous sodium chloride,

NaCl

, will not react with aqueous copper(II) sulfate,

CuSO

4

, because the two potential products are soluble in aqueous solution.

The chemical equation given to you is actually incorrect because copper(II) chloride,

CuCl

2

, is not insoluble in aqueous solution. In fact, it is quite soluble.

This means that the reaction does not produce an insoluble solid that precipitates out of solution.

Sodium chloride and copper(II) sulfate are both soluble ionic compounds that dissociate completely in aqueous solution

Explanation:

Korolek [52]4 years ago

5 0

Answer:

double replacement reaction

Explanation:

i took the test and got a 100

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Dvinal [7]

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4 years ago

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puteri [66]

Answer: The standard cell potential (E∘) for the reaction $X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$ is -0.121 V

Explanation:

The reaction is:

$X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = Standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

K= equilibrium constant = $8.97\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G^0=-(8.314J/Kmol\times 298K\times \ln (8.97\times 10^{-3})\\\\\Delta G^0=11678.9J/mol$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 1

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = ?

Putting values in above equation, we get:

$11678.9J/mol=-1\times 96500\times E^0_{cell}$

$\frac{11678.9J/mol}{-96500}=E^0_{cell}$

$-0.121V=E^0_{cell}$

Thus standard cell potential (E∘) for the reaction $X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$ is -0.121 V

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