UP PGT Chemistry Mock Test - 4 - UPTET MCQ

The expression for the change in internal energy is given by,
ΔU=q+w
q = heat absorbed = 200 J
w = work done = −P x V
=−2×10⁵ × 500×10⁻⁶
= −100 N - m
ΔU=200−100J= 100 J
Hence, the correct option is C.

Let's number our eqns-
CH₃OH + 3/2 O₂ → CO₂ + 2H₂O ...(1)
ΔH₁ = –726 kJ/mol
C + O₂ → CO₂ ...(2)
ΔH₂ = –393 kJ/mol
H₂ + 1/2 O₂ → H2O ...(3)
ΔH₃ = –286 kJ/mol
Eqn(2) + 2×eqn(3) - eqn(1) :-
C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂0→ CO₂ + 2H₂0 + CH₃OH + 3/2 O₂
C + 1/2 O₂ + 2H₂ → CH₃OH
Thus enthalpy of formation-
∆Hf(CH₃OH) = ∆H₂ + 2∆H₃ - ∆H₁
∆H_f(CH₃OH) = -393 + 2(-286) + 726
∆Hf(CH3OH) = -239 kJ/mol

The variability of oxidation state, a characteristic of transition element arises due to incomplete filling of d-orbitals.

4f electrons have greater shielding effect but 5f electrons have poor shielding effect.

The water which forms come with soap is known as hard water. The hardness that occurs due to the presence of bicarbonates of calcium and magnesium is temporary hardness. Temporary hardness can be removed by boiling and Clark’s process.

Elevation of boiling point and depression in freezing point are directly related to molality of solution.

(B) & (C) compounds are R-configuration.

RNO₂ +Fe/HCl ->RNH₂

Thus, AS behaves as a non-electrolyte at the freezing point of solution (remains non-ionised).

If CO₂ is passed through slaked lime(calcium hydroxide Ca(OH)₂), it gives out calcium carbonate (CaCO₃) also called marble/limestone.

CO₂ + Ca(OH)₂→ CaCO₃+ H₂O

On protonated epoxide, nucleophilic attack occur at α-carbon Where more stable carbocation could be formed.

Glycine is the simplest amino acid and is the only amino acid that is not optically active (it has no stereoisomers).

The correct answer is option  A
Reaction of propyne with 2 molecules of HBr to form 2,2 -dibromine propane can be written as:
CH₃ − C ≡ CH + 2HBr → CH₃​ − C(Br) = CH_2​
CH₃​ − C(Br) = CH_2​ + HBr → CH_3​ − C(Br)_2​ − CH_3​
The reaction follows Markownikoff's rule.

For geometry and shape of a molecule, both bond pair and lone pair of electron in valence shell of central atom.

The correct answer is option b. It has a center of symmetry: This statement is incorrect. A center of symmetry requires that for every atom at a certain position, there is an identical atom at an equal distance but opposite in direction. The given compound lacks a center of symmetry

Aggregation of colloidal particles into insoluble precipitate by addition of some suitable electrolyte is called coagulation.

 About 75% of the elements on the periodic table are metals. The only groups that are not metals are the noble gases, halogens, and the group actually called nonmetals.

Mass of O₂=32 gm ∆H/gm = -483.636/32 = -15.11
Mass of O₃=48 gm ∆H/gm = -868.2/48 = -18.08
Mass of H₂O₂=34 gm, ∆H/gm = -347.33/34 = -10.21
We can say that II has highest amount of energy per gram of oxidising agent.

The correct answer is option A
(I) ^xPO₄³⁻​ ⇒ x + 4 × (−2) = −3
                ⇒x = −3 + 8 = +5
                ⇒x = +5
Oxidation number of P = +5
(II) ^xSO₄²⁻​ ⇒ x + 4 × (−2) = −2
                ⇒x = −2 + 8
                  ⇒x = +6
Oxidation number of S=+6
(III) ^xCr₂​O₇²⁻ ​    ⇒2x + 7 × (−2) = −2
                        ⇒2x =−2+14
                        ⇒2x=12
                        ⇒x= 12/2​ = +6

Pentanedioic acid is glutaric acid.

For isobaric process, we have ∆S =nC_pln(T₂/T₁)
T2 = 273+127 = 400K and T₁ = 273+227 = 300K
Applying pV = nRT at initial condition,
1×10 = n×0.0821×300
n = 0.40
Applying ∆S =nC_pln(T₂/T₁)
∆S =0.40×5/2R×ln(400/300) = 2.38 JK^-1

DNA differs from RNA because DNA has deoxyribose sugar while RNA has ribose sugar. DNA has thymine as pyrimidine while, in RNA thymine is replaced by uracil.So, the correct answer is 'In the nature of sugar and pyrimidines'

For radius r,
circumference of the orbit = 2πr_n

Number of waves in one complete revolution = nλ
For third orbit = nλ = 3λ
Thus, three waves are formed in one revolution.