Chemistry Test 110

Result

Chemistry Test 110

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following causes laxative effect when present above 500 ppm in water :

A
NO₃^–

B
SO₄^{– –}

C
F^–

D
Zn⁺⁺
Question 2
4 / -1

Which is not a pair of reducing sugars-

A
Glucose, sucrose

B
Fructose, lactose

C
Glucose, maltose

D
Glucose, fructose
Question 3
4 / -1

Given the value of their van der Waals’ constant ‘a’ arrange the following gases in the order of their expected liquification pressure at a temperature T. T is below the critical point of all the

gases.

Gas CH₄ Kr N₂ Cl₂

‘a’(atmL².mol^–2) 2.283 2.439 1.408 6.579

A
CH₄ < Kr < N₂ < Cl₂

B
N₂ < CH₄ < Kr < Cl₂

C
Cl₂ < Kr < CH₄ < N₂

D
Cl₂ < N₂ < Kr < CH₄
Question 4
4 / -1

Which of the following d-block element has highest stable oxidation state.

A
+8 in Mn

B
+8 in Os

C
+8 in Cr

D
+8 in Fe
Question 5
4 / -1

Find D

A
(1)

B
(2)

C
(3)

D
(4)

Solution
Question 6
4 / -1

Calculate ΔS_universe for following chemical reaction

C (graphite) + 2H₂(g) → CH₄(g);

Δ_fH° = –74.81 kJ at 298 K.

The standard entropies of C (graphite), H₂(g) and CH₄(g) are 5.740, 130.684 and 186.264 J/K-mol, respectively.

A
170.1 J/K

B
125 J/K

C
212 J/K

D
138.2 J/K

Solution

ΔS_Reaction = [186.264] – [5.740 + 2 ×130.684)] or ΔS_System = – 80.844 J/K

ΔS_Surrounding = + ΔH/T = 74.81 × 10³/ 298 = 251 j/k

ΔS_Universe = – 80.844 + 251 = 170.1 J/K
Question 7
4 / -1

Some white colourless crystals are heated. A cracking sound is heard and brown fumes are given off and the residue left is yellow-brown in colour. When a glowing splinter is held in the fumes, it is relighted. The fumes consist of :

A
O₂

B
NO₂

C
Cl₂

D
NO₂ and O₂

Solution

Pb(NO₃)₂ PbO + 2 NO₂ + 1/2 O₂
Question 8
4 / -1

A
(1)

B
(2)

C
(3)

D
(4)

Solution

Vicinal diol compounds breakdown

by HIO₄.
Question 9
4 / -1

Which of the following structure is correctly drawn according to fundamental idea of VSEPR theory-

A
(1)

B
(2)

C
(3)

D
(4)

Solution

(A) θ incorrect less than 109º28'

(B) incorrect position of lone pair

(C) correct square planar structure with θ = 90º

(D) θ incorrect because it is square planar with two lone pair at Xe and θ = 90º
Question 10
4 / -1

The end products of following reaction would be,

A
(1)

B
(2)

C
(3)

D
(4)

Solution

3 moles of HCHO give cross aldol condensation with aromatic ketone one mole of HCHO under goes cross Cannizaro reaction with the aldol product.