UP PGT Chemistry Mock Test - 3 - UPTET MCQ

The reaction is as follow:-
Ca(HCO₃)₂(s)⇌CaO(s) + 2CO₂ (g) + H₂O(g)
At 25° C H₂O goes in liquid state
Kp = (PCaO)1×(PCO₂)₂
(PCa(HCO₃)₂)
Since, Ca(HCO₃)₂, CaO and H₂O are not in gaseous state, so their partial pressure is taken 1.
Putting all values, we have
36 = (PCO₂)₂ 
Or PCO₂ = 6 atm

[Hint: Reverse of (B) & (C) is spontaneous; weakest Oxidizing Agent here is Mg²⁺]

If FeCl₃ is added to excess of hot water, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.
Ferric hydroxide is a negative sol,FeCl₃ electrolyte will coagulate the most.

OF₂ Fluorine is most electronegative atom with oxidation number = - 1
x - 2 = 0
x = + 2
H₂O₂   2 + 2x = 0 ⇒ x = - 1 

∆p×∆x = h/4π
Or ∆v = h/4π∆xm      (since p = mv)
=6.6×10^-34Js/4×3.14×25×10^-3kg ×10^-5 
by solving this your answer is 2.1×10^-28

(a) 2s (b) 2 p (c) 3d (d) 3d
Thus, (c) and (d), both 3d, have equal energy.

The correct answer is option  C
Emulsifying agent
Emulsification is the process which involves the preparation of emulsion.Generally, an emulsion is prepared by subjecting a mixture of the immiscible liquid to a distinct layer upon standing. To prevent the separation of layers and to get the stable emulsion, a small quantity of the third substance is added. This substance which stabilizes the emulsion is called an emulsifier.
 

The correct answer is option A.
Negative deviation means lower vapour pressure, it suggests high boiling point, thus resultant intermolecular force should be stronger than individual one.

Valence bond theory was introduced by Heitler and London (1927) and further work was done by Pauling and others.

In the valence bond (VB) theory, proposed in large part by the American scientists Linus Pauling and John C. Slater, bonding is accounted for in terms of hybridized orbitals of the metal ion, which is assumed to possess a particular number of vacant orbitals available for coordinate bonding that equals its coordination number. 

Isobar means when pressure is constant.

2CO + O2 ------------> 2CO₂
Atomic wt of C =12 kg or gms, Atomic wt of O = 16 kg or gms
2(12 + 16) kg of CO + 32 kg of O2 -----------> 2(12+32) kg of CO₂
56 kg of CO + 32 kg of O₂ -------> 88 kg of CO₂
For, 1kg of CO + 32/56 kg of O₂ ---------> 88/56 kg of CO₂
Therefore, 1kg of CO requires 32/56 or 4/7 kg of oxygen to produce 88/56kg or 11/7 kg of CO2

On passing CO₂ gas, Ca(OH)₂ becomes CaCO₃ (milky white ppt.) which proves the presence of CO₂.

Most of the f block elements, lanthanoids and actinoids are man made/synthetic/prepared in the lab. Especially actinoids.

Energy of orbital is determined by n+l value. The orbital having higher n+l value will have higher energy. If n+l value is same then orbital having higher n value will have higher energy.

Atomic size decreases across the period.
Cl has a smaller size than Ar due to the inert nature atoms exist as single atoms.

For option a
12 = [Ne]3s2 , 20 = [Ar]4s2 , 4 = [He]2s2 , 88 = [Ra]7s2.
For option b
8 = [He]2s22p4 , 16 = [Ne]3s23p4 , 34 = [Ar]4s23d104p4 , 2 = 1s2
We can see that only in option a, the last electrons enter in similar group. So, option a is correct.
Similarly we can check for option c and d as done above. However they won't fall in the same group.

KCI is an ionic compound. Force of attraction (F) between K+


Greater the value of D, smaller the value of F, hence larger the solubility.
Note D for CH₃OCH₃ is least out of given solvents, thus solubility is minimum.

With increase in temperature molecular collisions increases therefore rate of reaction doubles.

Due to high lattice enthalpy of fluorides, it is difficult to break the lattice of fluoride compounds(due to the small size of fluorine, so the lattice is strong). SO the solubility of fluorides is less than chlorides.

Primary valency is equal to no. of ligands attachted which are negatively charges.

Oxidising power of F₂ > CI₂ > Br₂ > l₂


Thus, (II) is true.

In experiment (III), orange colour of Br₂ does not appear, as it also oxidises Kl to l₂. Hence , only violet colour appears in CHCI₃ layer. Thus, (III) is also true.

Answer:

Given

Density of water (ρ) = 1g/cm³

Molar mass of water (M) = 18 g/mol

∴ Molar volume (Vm) = M/ρ

Vm = 18 cm³/mol

∴ Volume of 1 molecule of water = V_m/N_A

Where N_A is Avogadro number

Volume of 1 molecule of water = 18/(6.022 × 10²³)

Volume of 1 molecule of water = 2.989 × 10^-23 ≈ 3 × 10^-23.

(C—Cl) bond has (C=Cl) double bond character, thus it is stable.

Benzyl carbocation is stable due to resonance, hence (C—Cl) bond is cleaved easily.

• Partition chromatography because the substances are partitioned or distributed between liquid phases. 
• The two phases are water held in pores of the filter paper and the other phase is a mobile phase that passes through the paper.

(V) form s m ost stable, tertiary allylic, carbocation. (II) is vinylic halide, least reactive.

The number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃ is four. This is calculated by the formula 2ⁿ⁺¹. where n is number of chiral centres, which is 1 in this case. So, 2¹⁺¹ = 2²= 4

Hydrogen peroxide, represented by H₂O₂, is stored in acetanilide or sodium pyrophosphate, the presence of traces of alcohol, which slows down the rate of decomposition of hydrogen peroxide (H₂O₂).

 (d) Aqueous 6% methanol solution by mass means 100 g of aqueous solution has = 6g CH₃OH