ZnSO 4- 7 H 2 O Below 70- C Above 39- C - A Above 70- C - B Above 280- C - O 2- SO 2- ZnO - A and B respectively are-A- ZnSO 4- H 2 O - ZnSO 4- 6 H 2 OB- ZnSO 4 -2 H 2 O - ZnSO 4 -6 H 2 OC- ZnSO 4- 6 H 2 O - ZnSO 4- 2 H 2 OD- ZnSO 4- 6 H 2 O - ZnSO 4- H 2 O

The correct option is D ZnSO_4.6H_2O & ZnSO_4.H_2OZnSO_4.7H_2O ZnSO_4.6H_2O ZnSO_4.H_2O O_2 +SO_2 +ZnO ZnSO_4 (Anhydrous) This reaction is self ex ...

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Byju's Answer

Standard IX

Chemistry

Zinc II Sulphate

ZnSO 4· 7 H 2...

Question

$Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C A A b o v e 70^{\circ} C - ------ \to B A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O$ . A and B respectively are:

Z n S O_{4} . H_{2} O & Z n S O_{4} .6 H_{2} O

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Z n S O_{4} .2 H_{2} O & Z n S O_{4} .6 H_{2} O

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Z n S O_{4} .6 H_{2} O & Z n S O_{4} .2 H_{2} O

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Z n S O_{4} .6 H_{2} O & Z n S O_{4} . H_{2} O

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Solution

The correct option is D $Z n S O_{4} .6 H_{2} O & Z n S O_{4} . H_{2} O$
$Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C Z n S O_{4} .6 H_{2} O A b o v e 70^{\circ} C - ------ \to Z n S O_{4} . H_{2} O A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O 800^{\circ} C \leftarrow -- - \begin{matrix} Z n S O_{4} (A n h y d r o u s) \end{matrix}$

This reaction is self-explanatory. As $Z n S O_{4} .7 H_{2} O$ is heated above 39 degree Celsius and below 70 degree Celsius, it loses one water molecule and further when $Z n S O_{4} .6 H_{2} O$ is heated above 70 degree Celsius, it loses 5 water molecules. Therefore, A is $Z n S O_{4} .6 H_{2} O$ and B is $Z n S O_{4} . H_{2} O$ .

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Similar questions

Q. Consider the following reactions,
(I)

Z n + d i l . H_{2} S O_{4} \to Z n S O_{4} + H_{2}

(II)

Z n + c o n c . H_{2} S O_{4} \to Z n S O_{4} + S O_{2} + H_{2} O

Oxidising agents in

I

and

I I

are _______ respectively

Q. Consider the following unbalanced reactions:

I : Z n + d i l . H_{2} S O_{4} ⟶ Z n S O_{4} + H_{2}

I I : Z n + c o n c . H_{2} S O_{4} ⟶ Z n S O_{4} + S O_{2} + H_{2} O

I I I : Z n + H N O_{3} ⟶ Z n {({N O}_{3})}_{2} + {N H}_{4} {N O}_{3} + H_{2} O

Oxidation number of hydrogen changes in :

Q. What kind of reaction is this ?

Z n O + H_{2} S O_{4} \to Z n S O_{4} + H_{2} O

The major products of the following reaction

$Z n S (s) + O_{2} (g) h e a t - --- \to$

Are

Q. State the type of reaction given below.

{Z n O}_{(s)} + {H_{2} S O_{4}}_{(a q)} \to {Z n S O_{4}}_{(s)} + {H_{2} O}_{(a q)}

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ZnSO4.7H2OAbove 39∘C−−−−−−−→Below 70∘CAAbove 70∘C−−−−−−−→BAbove 280∘C−−−−−−−→O2+SO2+ZnO. A and B respectively are:

$Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C A A b o v e 70^{\circ} C - ------ \to B A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O$ . A and B respectively are: