CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Oxidising Reagent and Reducing Reagent

The equivalen...

Question

The equivalent weight of the underlined compound by taking its molecular weight as $M$ is:

$N H_{4} {N O}_{3} - --------- - ⟶ N_{2} O + 2 H_{2} O$

A

E = \frac{M}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

E = \frac{M}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

E = \frac{M}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

None of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D None of these
$\begin{matrix} N H_{4} N O_{3} ⟶ N_{2} O + 2 H_{2} O Change in Oxidation number of Nitrogen = 1 Equivalent weight = M \end{matrix}$ (+5 to +4)

Suggest Corrections

thumbs-up

0

Similar questions

4 {F e}^{2 +} + 2 {N H}_{2} O H - --------- - ⟶ 4 {F e}^{3 +} + N_{2} O + 4 H^{+}

The equivalent weight of the underlined compound by taking its molecular weight as

M

P - - {C l}_{5} ⟶ P {C l}_{3} + {C l}_{2}

The equivalent weight of the underlined compound by taking its molecular weight as

M

H_{2} C_{2} O_{4} - ------- - + M n O_{2} + 2 H^{+} ⟶ {M n}^{2 +} + 2 H_{2} O + 2 C O_{2}

The equivalent weight of the underlined species by taking its molecular weight as

M

2 {H S}^{-} - ----- - + 4 H S O_{3}^{-} ⟶ 3 S_{2} O_{3}^{2 -} + 3 H_{2} O

The equivalent weight of the underlined compound by taking its molecular weight as

M

3 {C l}_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} - ------- - + 3 H_{2} O

The equivalent weight of the underlined compound by taking its molecular weight as

M

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Rules for Assigning Oxidation Number and State

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Oxidising Reagent and Reducing Reagent

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon