Chemistry Test 253

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Chemistry Test 253

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Weekly Quiz Competition

Question 1
4 / -1

For the following electrochemical cell reaction at 298 K:

Zn(s) + Cu²⁺(aq) ⇌ Cu(s) + Zn²⁺(aq),

Given that the standard cell potential (E° cell) is +1.10 V.

A
0.027

B
37.22

C
0.012

D
85.73

Solution
Question 2
4 / -1

We know that the relationship between K_c and K_p is K_p = K_c (RT)^ΔnWhat would be the value of Δn for the reaction NH₄Cl (s) ⇔ NH₃ (g) + HCl (g)

A
1

B
0.5

C
1.5

D
2

Solution
Question 3
4 / -1

The pH of a 10⁻⁸ molar solution of HCl in water is

A
8

B
between 7&8

C
between 6&7

D
None of these

Solution

pH = −log(H⁺) = −log10⁻⁸= 8

As the solution is acidic, pH<7. This is because [H⁺]from H₂₀[10⁻⁷] cannot be neglected in comparison to 10⁻⁸.

It is not possible for acid, so it is [H⁺], the [H⁺]of water is also added

Total [H⁺] in solution

= [H⁺] of HCl⁺[H⁺] of water

= (1 × 10⁻⁸+ 1 × 10⁻⁷)M

= (1 + 10) × 10⁻⁸= 11 × 10⁻⁸M

∴ pH = −log[H⁺] = −log11 × 10⁻⁸

= −log11 + 8log10

= −1.0414 + 8

= 6.9586, between 6 and 7.
Question 4
4 / -1

Ag₂CrO₄ and Ag₂CO₃ are simultaneously dissolved in water. The solubility of Ag₂CrO₄& Ag₂CO₃ are 'x′ and 'y′mol/l respectively. If the Ksp of Ag₂CrO₄ is 'a′ and that of Ag₂CO₃ is 'b', then which of the following is correct?

A
(x + y)2x = a

B
(2x + 2y)y = b

C
(2x + 2y)²x = b

D
(2y + 2x)²y = b

Solution
Question 5
4 / -1

The straight chain polymer is formed by:

A
hydrolysis of CH₃SiCl₃ followed by condensation polymerisation

B
hydrolysis of (CH₃)₄ Si by addition polymerisation

C
hydrolysis of (CH₃)₂SiCl₂ followed by condensation polymerisation

D
hydrolysis of (CH₃)₃SiCl followed by condensation polymerisation

Solution

Hydrolysis of substituted chlorosilanes yield corresponding silanols which undergo polymerisation.

Polymerisation of dialkyl silandiol yields linear thermoplastic polymer.
Question 6
4 / -1

Calculate the boiling point of a one molar aqueous solution (density = 1.04 gmL⁻¹) of potassium chloride, K_b for water = 0.52 kg mol⁻¹. Atomic mass of K = 39,Cl = 35.5

A
107.28°C

B
103.68°C

C
101.078°C

D
None of these

Solution

Volume of solution = 1000 ml

Mass of the solution = V × d

= 1000 mL × 1.04 g/mL = 1040 g

Amount of solute in 1000 mL solution

= 1 M

= 1 M × Molecular Mass = 1M × 74.5 g/mol

= 74.5 g

Mass of water = Mass of solution − Mass of

KCl = 1040 g−74.5 g = 965.5 g

Molality of the solution = m =

= 1.0357

ΔT_bi.K_b⋅m = 2 × 0.52 K kg mol⁻¹× 1.0357

= 1.078°C.

Boiling point of the solution

=100^∘C + 1.078^∘C = 101.078^∘C
Question 7
4 / -1

Ammonia reacts with oxygen giving nitrogen and water. If the rate of formation of N₂ is 0.70 mol L⁻¹ s⁻¹, the rate at which O₂ is consumed is

A
1.05 mol L⁻¹s⁻¹

B
0.70 mol L⁻¹ s⁻¹

C
2.10 mol L⁻¹s⁻¹

D
0.35 mol L⁻¹ s⁻¹

Solution
Question 8
4 / -1

The half-life for the thermal decomposition of acetone is 80 s and is independent of initial concentration of acetone. The time required for the reaction to go to 80% decomposition is (Given: log2 = 0.30)

A
186.1 s

B
206.1 s

C
150.1 s

D
226.1 s

Solution

The reaction follows first-order kinetics. Hence k = 0.693/_t1/2= 0.693/(80 s). For the reaction to be 80%, [A]_t/[A]₀= 0.2. Hence
Question 9
4 / -1

The correct order of acidic strength is

A
Cl₂O 7 > SO₃> P₄O₁₀

B
CO₂> N₂O₄> SO₃

C
Na₂O > MgO > Al₂O₃

D
K₂O > CaO > MgO

Solution

The acidic strength of an acid is defined as the tendency of the acid to dissociate into its corresponding ions in order to release a proton and the anion. The acidic strength of a strong acid is high and that of a weak acid is low. This means that on dissociation, a strong acid releases its proton at a much faster rate than the weak acid.

Oxidation number: The greater the oxidation number of the central atom of the oxide, the greater the acidic strength will be.

n a periodic table, the acidic strength of an atom increases across a period and decreases down the group. In the above options, the correct order of acidic strength of the oxides is:

Cl₂O₇> SO₃> P₄O₁₀

The oxidation state of the oxides is as follows:

Let the oxidation number of the chlorine atom be 'x′.

Cl₂O₇= 2x + 7(−2) → x = +7

Let the oxidation number of the sulfur atom be ′y′.

SO₃= y + 3(−2) ⇒ y = +6

Let the oxidation number of the phosphorus atom be ′z′.

P₄O₁₀− 4z + 10(−2) → z − +5

Thus, as we can see that the order of oxidation is in a decreasing manner, this means that this option is correct.

So, the correct answer is Option (a).
Question 10
4 / -1

What are X and Y in the following reactions?

A

B

C

D

Solution

(i) Since H₂O is highly polar in nature and aniline is highly reactive towards electrophilic substitution.

Thus, Br₂ in H₂O gives: 2, 4, 6 tribromo aniline as main product i.e.

and as a result only mono-substituted bromine at p-position is the major product i.e.,