AWES PGT Chemistry Mock Test - 7 - AWES TGT/PGT MCQ

The element of Copernicium’s atomic number is 112. it is otherwise called as Ununbium as per the IUPAC nomenclature. The name Copernicium is given after the astronomer Nicolaus Copernicus. It’s a d-block transactinide element that belongs to Group-12.

Cl₂(g) is formed at anode and NaOH and H₂ are formed at cathode. Since, cathode and anode are separated. Hence. there is no reaction between the products formed at the cathode and anode. 

Note: but these reaction do not occur

Note: NaCl is neutral pH = 7 initially. After electrolysis , NaOH is formed

Hence, solution become, basic pH >7

(d) Among alkali metals, melting point decreases as the strength of metallic bonding decreases with increasing size of the atom. Thus, Cs has the lowest melting point (28.5°C) and will melt at 30°C.

The substance that gets adsorbed on the surface is known as adsorbate.

Zn + 2HCl ------> ZnCl₂ + H₂ since it forms 2 moles of hydrogen it is a dihydrogen.

CaNCN + 3H₂O → CaCO₃ + 2NH₃

Buna N is used in making oil seals , tank linings.

In Benzene ring, the meta position is more stable than ortho and para.
At high temperature meta position will exist 
i.e option B is correct
 

From ΔT, of cane sugar, K, can be determined. 

For NaCI solution,

Green flame is given by Ba²⁺

BaS₂O₃+K₂CrO₄ → BaCrO₄+K₂S₂O₃   yellow ppt

The modern periodic table orders elements by increasing atomic number, which is the number of protons in the atom of the element. Mendeleev did not know about the parts of the atom, so he used the next best thing -- atomic weight.

The correct answer is option C
ΔT = Kf​ × molality × (1+α)(HX ⇌ H+   + X−)
Given molality = 0.2;  degree of ionization is 0.3
=1.85×0.2×(1+0.3)
=0.481K
The freezing point of the solution will be 0.81 K.

Inversion of configuration takes place at chiral carbon.

The Correct answer is option B
Hoffman bromamide degradation reaction is shown by ArCONH₂​.
Where the aryl amide is converted to aryl amine in the presence of Br₂​ and NaOH

Zero overlap means that the orbitals don’t overlap at all. When there is no overlapping the bond formation doesn’t occur. As we all know that the extent of overlapping is dependent on the strength of the bond.

C₆H₅N(CH,)₂ this is tertiary amine.

(II) forms more stable carbocation than (I).

The major product is primary amine.

The catalytic reaction that depends upon the structure of pores of the catalyst and the size of the reactant and product molecules is called shape/selective catalysis. Zeolites are good shape selective catalysts because of their honeycomb structure. 

Thermal stability decreases as the strength of H-X bond decreases which in turn decreases as the size of the halogen increases, i.e. HF > HCI > HBr > HI

200 g H₂O at 60°C has 95 g NH₄Cl

Based on solubility graph, at 20°C,

100 g of H₂O can dissolve = 36 g NH₄CI

hence, 200 g of H₂0 can dissolve = 72g  NH₄CI

Thus, NH₄CI separated (crystallised) = 95 - 72 = 23 g

Mass % = 36%

1.38 moles of CH₃OH in w kg water

1.38 moles of CH₃OH in w Kg water 

So that molarity = 1.0

Cesium is an alkali metal, like sodium; hence, it forms a salt, cesium chloride (CsCl) similar to table salt, sodium chloride.CsCl looks like most salts:It's worth noting that the formation of salts is highly exothermic, that is, if you were to get cesium metal and gaseous chlorine to combine to form this salt, the reaction would produce a lot of heat along the way!

Cu⁺ = [Ar] 3d¹⁰, Cu⁺ has inner 18-electrons shell. Shielding effect is decreased. Thus, electronegativity is increased. This results in greater polarising power of Cu⁺ .
Hence, CuCI is covalent.

According to the electronic configuration of sodium(Na),it has 1 valence electron(2,8,1).So it loses one electron to form a cation and becomes sodium cation (Na+).As it loses one electron ,the outermost shell has 0 electrons (i.e it gets eliminated ) and Na+ has only 2 shells. Hence, it lost one of its shell, the sodium ion is smaller than the sodium atom.

H is reduced to H₂ and O is oxidised to O₂.

Nucleophilic attack in step-ll occur at the carbon atom which can better accommodate the positive charge. Hence, attack of Br^- or Cl^- in second step occur at 2° carbon rather that at 1° carbon.

Black P has layer lattice like graphite.