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3 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]3 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]3 years ago

5 0

Answer:

double replacement reaction

Explanation:

i took the test and got a 100

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3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 m

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Answer:

$49.45~%$

Explanation:

In this case, we have to start with the <u>chemical reaction</u>:

$C_6H_1_2O~->~C_6H_1_0~+~H_2O$

So, if we start with <u>10 mol of cyclohexanol</u> ( $C_6H_1_2O$ ) we will obtain 10 mol of cyclohexanol ( $C_6H_1_0$ ). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

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With this value we can calculate the grams:

$10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0$

Now, we have as a product 500 mL of $C_6H_1_0$ . If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

$500~mL\frac{0.811~g}{1~mL}=405.5~g$

Finally, with these values we can calculate the <u>yield</u>:

$%~=~\frac{405.5}{820}x100~=~49.45%$ %= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

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