Which of the following group is NOT Dobereiner Tri | Chemistry Questions & Answers

Q:

The reduction potential of hydrogen half-cell will be negative if :
(1) $P (H_{2})$ = 2 atm and [H+] = 2.0 M
(2) $P (H_{2})$ = 1 atm and [H+] = 2.0 M
(3) $P (H_{2})$ = 1 atm and [H+] = 1.0 M
(4) $P (H_{2})$ = 2 atm and [H+] = 1.0 M

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: D) Option 4

Explanation:

$H^{+} - e^{-} \to \frac{1}{2} H_{2}$

Apply Nernst equation

$E = 0 - \frac{0.059}{1} \log (\frac{P H_{2}^{\frac{1}{2}}}{[H^{+}]})$

$E = - \frac{0.059}{1} \log (\frac{2^{\frac{1}{2}}}{1})$

Therefore E is Negative

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Q:

Trichloroacetaldehyde was subject to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate and another compound. The other compound is :

A) Chloroform	B) 2, 2, 2-Trichloroethanol
C) Trichloromethanol	D) 2, 2, 2-Trichloropropanol

Answer & Explanation Answer: B) 2, 2, 2-Trichloroethanol

Explanation:

$\begin{matrix} C l \\ | \\ C l & - & C & - & C H O \\ | \\ C l \end{matrix}$ $\to_{C a n n i z z a r o' s R e a c t i o n}^{N a O H}$ $\begin{matrix} C l \\ | \\ C l & - & C & - & C O O N a \\ | \\ C l \end{matrix}$ $+$ $\begin{matrix} C l \\ | \\ C l & - & C & - & C H_{2} \\ | 2 & 1 \\ C l \end{matrix}$ $- O H$

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7 5217

Q:

Silver Mirror test is given by which one of the following compounds?

1. Benzophenone

2. Acetaldehyde

3. Acetone

4. Formaldehyde

A) Only 1	B) Only 2
C) 1 and 3	D) 2 and 4

Answer & Explanation Answer: D) 2 and 4

Explanation:

Both Formaldehyde and Acetaldehyde will give this test

$H C H O \overset{[A g (N H_{3})_{2}^{+}]}{\to} \overset{A g}{S i l v e r m i r r o r} + O r g a n i c c o m p o u n d$

$C H_{3} - C H O \overset{[A g (N H_{3})_{2}^{+}]}{\to} \underset{S i l v e r m i r r o r}{A g} + O r g a n i c c o m p o u n d$

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66 63453

Q:

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 $d m^{3}$ at 27°C is to a volume of 100 $d m^{3}$

A) 42.3 J/ mole / K	B) 38.3 J/ mole / K
C) 35.8 J/ mole / K	D) 32.3 J/ mole / K

Answer & Explanation Answer: B) 38.3 J/ mole / K

Explanation:

$∆ s = n R \ln \frac{v 2}{v 1}$

$= 2.303 n R \log \frac{v 2}{v 1}$ = $2.303 \times 2 \times 8.314 \times \log (\frac{100}{10})$ = $38.3 J$ $m o l^{- 1}$ $k^{- 1}$

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43 16188

Q:

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

A) 518 nm	B) 1035 nm
C) 325 nm	D) 743 nm

Answer & Explanation Answer: D) 743 nm

Explanation:

$\frac{1}{λ} = \frac{1}{λ 1} + \frac{1}{λ 2}$

$\frac{1}{355} = \frac{1}{680} + \frac{1}{λ 2}$

$\frac{1}{λ 2} = \frac{680 - 355}{680 \times 355}$

$\Rightarrow λ 2 = 743 n m$ .

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Q:

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is

1) $A_{2} B_{5}$

2) $A_{3} B_{2}$

3) $A_{2} B_{3}$

4) $A_{3} B_{4}$

A) Option 1	B) Option 2
C) Option 3	D) Option 4

Answer & Explanation Answer: A) Option 1

Explanation:

$Z_{A} = \frac{8}{8}$ ; $Z_{B} = \frac{5}{2}$

So formula of compound is $A B_{\frac{5}{2}}$

i.e., $A_{2} B_{5}$ .

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13 9141

Q:

The degree of dissociation ( $α$ ) of a weak electrolyte, $A_{x} B_{y}$ is related to van't Hoff factor (i) by the expression:

$1 . α = \frac{x + y + 1}{i - 1}$

$2 . α = \frac{x + y - 1}{i - 1}$

$3 . α = \frac{i - 1}{x + y + 1}$

$4 . α = \frac{i - 1}{x + y - 1}$

A) 1 is correct	B) 2 is correct
C) 3 is correct	D) 4 is correct

Answer & Explanation Answer: D) 4 is correct

Explanation:

Van't Hoff factor (i) = $\frac{O b s e r v e d C o l l i g a t i v e P r o p e r t y}{N o r m a l C o l l i g a t i v e P r o p e r t y}$

$\begin{matrix} A_{x} B_{y} \\ 1 - α \end{matrix} \to \begin{matrix} x A^{+ y} \\ x α \end{matrix} + \begin{matrix} y B^{- x} \\ y α \end{matrix}$

Total Moles = $1 - α + x α + y α$ = $1 - α (x + y - 1)$

$i = \frac{1 + α (x + y - 1)}{1}$

$\Rightarrow α = \frac{i - 1}{x + y - 1}$

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6 7263

Q:

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of :

A) An acetylenic triple bond	B) Two ethylenic double bonds
C) A vinyl group	D) An isopropyl group

Answer & Explanation Answer: C) A vinyl group

Explanation:

$C =} C H_{2} \overset{O z o n o l y s i s}{\to} H C H O$

Vinylic group.

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3 4816

A) Li, Na, K	B) N, P Au
C) Cl, Br, I	D) Ca, Sr, Ba

Which of the following group is NOT Dobereiner Triads?

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