HPSC PGT Chemistry Mock Test - 4 - HPSC TGT/PGT MCQ

Key Points

Rock cycle:​​

• Rocks do not remain in their original form for a long period.
• They undergo a transformation.
• This cycle is an uninterrupted process through which old rocks are converted into new ones.
• Igneous rocks are primary rocks.
• These rocks can be changed into metamorphic rocks.
• Sedimentary and metamorphic rocks are formed from these primary rocks.
• The fragments evolved out of metamorphic rocks and igneous again form into sedimentary rocks.
• Sedimentary rocks themselves can develop into fragments.
• The crustal rocks -igneous, metamorphic, and sedimentary-once formed may be carried down into the interior of the Earth through subduction.
• In this process, parts or entire crustal plates subduct under another plate and the same melt at high temperatures in the interior.
• This results in the formation of molten magma, the unique source for igneous rocks.
• The process of transformation follows the steps,  'Igneous rocks - sediment - sedimentary rocks - metamorphic rocks - magma'​ 
• The image below clearly explains the rock cycle.

Hence, rock cycle is the process of transformation of rock from one to another

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

Solubility of an ionic compound in water depends upon lattice energy of compound and hydration energy of ions.

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.

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

Clemmensen reduction is suitable for reductio n of carbonyls containing additional acidic functional group.

Carbon has 4 electron in its outermost shell. So it is difficult to loose electrons or gain electrons. That's why carbon forms covalent bond by sharing of electrons.

∆G° = ∆H° - T∆S°
       = -54070 - 298 10
       = -57050
∆G° = -2.303 RT log₁₀k
-57050 = -2.303 8.314 298 log₁₀k
57050 = 5705 log₁₀k
log₁₀k = 10

If free radical halogenation generate a chiral carbon, racemic mixture of halides are always formed due to equal probability of halogenation from both sides of planar free radical.

The correct answer is Option D
 When alum is added to raw water it reacts with the bicarbonates alkalinities present in water and forms a gelatinous precipitate.
It neutralizes all the suspended impurities of water resulting in their coagulation.

 If one considers following molecule, it is 

Symmetrical and thus
In the given species

BH₄– has 4 bond pairs and 0 lone pairs, so it is tetrahedral. CO₃(2^-) has 3 bond pairs and 0 lone pairs, so it is trigonal planar. NH₂- has 4 bond pairs and 2 lone pairs, so it is bent or angular. H₃O^+​ has 4 bond pairs and 1 lone pair, so it is a trigonal pyramid.

The correct answer is option A
Compare straight line equation
(y = mx + c) with Kt = -log (Ct) + log(C0)
where,
K being constant,
"t" is equivalent to "y" in straight line equation
(-1) is slope for equation which is "m" in straight line equation
log (Ct) is x-component
& log (C0) is y-intercept.
 

For second order reaction half life is inversely related to concentration of reactant.

he most stable conformation of the molecule shown (with two iodine and two carboxylic acid groups attached to a central carbon) would be the anti conformation. In this conformation, the bulky groups (such as the iodine atoms and the carboxylic acid groups) are positioned opposite to each other to minimize steric repulsion. This arrangement reduces the overall energy of the molecule, making it the most stable configuration.

Therefore, for the structure shown, arranging the I and COOH groups opposite each other would yield the most stable anti-conformation.

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 B
Given,
λCo(NH3)4Cl2+=50       λClo-4 =70
λ∞= λCo(NH3)4Cl2+  + λClo-4
λ∞= 50             + 70
λ∞=120
(x) Cell constant = 1/A
0.02 = l/A
Resistance(R)   =33.5Ω
K =c.x              (x = is cell constant)

S  =49.7 mol/L

Cl, O, and N forms Cl₂, O₂ and N₂ respectively to satisfy octet rule but He has noble gas configuration so do not form any compound and exist as monoatomic gas.

7-Bromocycloheptane-1,3,5-triene

Stoichiometric compounds of hydrogen are formed with s-block elements. Only Group I elements form stoichiometric hydrides with Hydrogen as same as halides.

Presence of either plane of symmetrical or centre of symmetry. Optical activity of one half of the molecule is exactely cancelled by other half and is internally compensated.

y = 3 = number of ions from one unit

i = van't Hoff factor = 1 + (y - 1)x = (1 + 2x)

Also, 10 mL of dibasic acid = 12mL of 0.1 N NaOH

N₁ (dibasic acid) = 0.12 g equivL^-1 

0.12 x equivalent weight = concentration = 10 g L^-1'

[Kr]4d¹⁰5s⁰

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

Cis-trans Isomers:- It is under the class of geometric isomer,

• It exists in alkenes (organic molecules which have double bonds). 
• When two similar or higher priority groups are attached to the carbon on the same side, it is termed as cis isomer and when it is attached to the opposite side it is called a trans isomer.

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