CrO3-H2O- A BIdentify A and B-

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Byju's Answer

Standard XII

Chemistry

Potassium Dichromate

CrO3+H2O⟶ A B...

Question

$C r O_{3} + H_{2} O ⟶ A {O H}^{-} - -- \to B$
Identify $A$ and $B$ ?

H_{2} C r O_{4}, H_{2} C r O_{7}

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H_{2} C r O_{7}, {C r}_{2} O_{3}

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C r O_{4}^{2 -}, {C r}_{2} O_{7}^{2 -}

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H_{2} C r O_{7}, C r O_{4}^{-}

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Solution

The correct option is D $H_{2} C r O_{7}, C r O_{4}^{-}$
$C r O_{3} + H_{2} O ⟶ H_{2} C r O_{7}$ (chromic acid)
$C r O_{7}^{2 -} + O H^{-} \to C r O_{4}^{-}$

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Similar questions

Q. Assertion :

{C O}_{2}

containing

H_{2} O

changes yellow

{C r O}_{4}^{2 -}

into

{C r}_{2} O_{7}^{2 -}

. Reason: Aqueous

{C O}_{2}

is acidic

(H_{2} {C O}_{3} ⇌ H^{+} + {H C O}_{3}^{-})

which converts

{C r O}_{4}^{2 -}

into

{C r}_{2} O_{7}^{2 -}

Balance the following redox reactions in acidic solutions:

U O_{2} + {C r}_{2} O_{7}^{2 -} ⟶ U O_{2}^{2 +} + {C r}^{2 +} + H_{2} O

Q. In the reaction,

{({N H}_{4})}_{2} {C r}_{2} O_{7} Δ - \to N_{2} + H_{2} O + {C r}_{2} O_{3}

The coefficient of

H_{2} O

is :

A(orange)

Δ - \to C (g) + r e s i d u e (c) (g r e e n) + H_{2} O

↓ N a O H

D(g)

H C l - - \to

white fumes

B(g)

A l - \to Δ (E) H_{2} O - - \to g a s (B)

Residue (C)

c o n c . H C l - ---- \to (F) N a O h, H_{2} O_{2} - ------ \to Δ G

(yellow)

C H_{3} C O O H - -------- \to (C H_{3} C O O)_{2} P b

H (yellow

If true enter 1, else enter 0.

Compounda (A) to (H) are identified as (A)

({N H}_{4})_{2} {C r}_{2} O_{7}

, (B)

N_{2}

, (C)

{C r}_{2} O_{3}

, (D)

{N H}_{3}

, (E)

A I N

, (F)

{C r C l}_{3}

, (G)

{N a}_{2} {C r O}_{4}

and (H)

{P b C r O}_{4}

{C r}_{2} O_{7}^{2 -} H^{+} - - \to C r

. Eq.wt of

{C r}_{2} O_{7}^{2 -}

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CrO3+H2O⟶AOH−−−−→BIdentify A and B?

The correct option is D H2CrO7,CrO−4CrO3+H2O⟶H2CrO7 (chromic acid)CrO2−7+OH−→CrO−4

$C r O_{3} + H_{2} O ⟶ A {O H}^{-} - -- \to B$
Identify $A$ and $B$ ?

The correct option is D $H_{2} C r O_{7}, C r O_{4}^{-}$
$C r O_{3} + H_{2} O ⟶ H_{2} C r O_{7}$ (chromic acid)
$C r O_{7}^{2 -} + O H^{-} \to C r O_{4}^{-}$