Chemistry (Engineering Entrance)

When oxygen oxidizes some of the metal ions and this increases the number of positive centres, hence the conductivity increases.

When sugar is added to water and a mixture of these two is formed, it suffers from lowering in freezing point and the freezing point of water is lowered and hence the first solid which separates out is ice. It is one of the colligative properties of a solution which is subjected to a change by the addition of solute.

 
We know that low pressure favours the reaction accompanied with increase in number of moles and in the given option, the number of moles of reactants are less than number of moles of product or we can say there is an increase in number of moles in product side, so this reaction will be favoured at low pressure.

 
We know that acidic strength increases with increase in oxidation number.Here we have oxidation number of chlorine as follows:<o:p>
HClO = +1, HClO₂ = +3, HClO₃ = +5, HClO₄ = +7. <o:p>In the given option the order given is correct.

Fusion bomb involves the principle pf nuclear fusion according to which when two lighter nuclei combine to form a bigger nucleius, then it is characterised by emission or liberation of tremendous amount of energy.

The electronic configuration of sulphur is 1s²2s²2p⁶3s²3p⁴ and oxygen is  1s²2s²2p⁴ , in terms of shell configuration it can be represented as follows: S=2,8,6 and O=2,6. Since the valence shell is having deficiency of two electrons and hence possesses, high electron affinity. Now when sulphur and oxygen both will accept one electron then their electronic shell configuration will become S = 2,8,7 and O = 2,7. Now it is easy for sulphur to accept one more electron as compared to oxygen because it is having three shells and hence the nuclear attraction in valence shell will be less as compared to oxygen having only two shells, hence the second electron affinity of sulphur will be less than that of oxygen.

Chemisorption is irreversible in nature as it involves chemical changes and we know that chemical changes are irreversible.

 
Cyclic trimer of sulphur trioxide is (SO₃)₃ , the oxidation of sulphur in it can be calculated as follows:<o:p>
Let the oxidation state of sulphur be x,hence 3x + 18 = 0, so x = 18/3 = +6

Since all the species given in the option are isoelectronic and we know that for isoelectronic species, the ionic radii increases with decrease in nuclear charge and in the given  question the order of decrease of nuclear charge is  N^3- > O^2- > F^- > Na⁺ which is the same as given in the option .

Glucose is a covalent compound, hence it cannot give ions in solution and hence will be possessing the least precipitating power.

In C₆H₅NO_2, the electron withdrawing group -NO₂ deactivates the benzene ring to the extent that it cannot respond to Friedel Craft reaction. Hence, this option is correct.

In this option statement 1 is said to be incorrect which is true as alkali metals get easily oxidized and not reduced by donating electrons from outermost shell and statement 2 is said to be correct which is true as alkali metals possess low value of ionization enthalpy.

 If the lattice energy is less than the hydration energy,  more heat is needed to be supplied in order to dissolve the salt in the solution.Hence it will be exothermic.

Statement 1 is correct as charge to mass ratio of the particles of anode rays depends upon the nature of the gas as it is not emitted from anode , statement 2 is also correct that anode rays particles consist of positive ions , and statement 2 is not the explanation of statement 1.

Barium sulphate is water soluble and opaque to X-rays,  hence it can be used as barium metal for getting  X-ray spectrum of the human digestive system.

We know that as per the molecular orbital theory,O₂ contains two electrons in the anti-bonding orbital if we have a look on the molecular orbital diagram of O₂ and hence, it is paramagnetic in nature. Also statement 2 is correct that O₂ contains two unpaired electrons in anti bonding orbital, and statement 2 is correct explanation of statement 1 as discussed above. 

 
Let the number of oxides be = x<o:p>
So number of octahedral  voids = x<o:p>
And number of tetrahedral voids = 2x<o:p>
Number of A²⁺ ion = 2x/8 = x/4<o:p>
Number of B³⁺ ions = x/2<o:p>
A²⁺ : B³⁺ : O^2-<o:p>
x/4: x/2: x<o:p>
1:2:4 <o:p>Hence the formula for the oxide is AB₂O₄

Monoamino dicarboxylic acids, have one extra -COOH and hence more base is required to neutralize. Therefore, pH will be lower than 7.

 It is the extensive cross-linking of the thermosetting polymer that it becomes infusible on heating.

 
Applying the Rydberg's equation 1/λ = R₀[1/n₁² - 1/n₂²] . <o:p>We get 1/λ = R₀ [3/4] or λ = 1.33/R₀. This value of wavelength is the least among the given options.

Given statement regarding the liquid junction potential is incorrect as salt bridge is used to reduce the liquid junction potential.

 
Pseudo alum has the general formula M^IISO₄. M₂^III (SO₄)₃. 24H₂O. <o:p>So on  comparing we get FeSO₄.Al₂(SO₄)₃.24H₂O is pseudo alum.

Schotten-Baumann reaction is the acyclation of alcohols, phenols and amines, and not ethers, by shaking with aromatic acyl chloride in the presence of a base, usually aqueous sodium hydroxide. Hence, this test is not given by ethers.

 Plumbo-solvency is said to take place when Pb gets converted to its hydroxide on reaction with water in presence of oxygen,  the given option satisfies all the conditions. 

The answer can be calculated as follows:<o:p>
From observation 1, we observe that reaction is 1^st order with respect to A. <o:p>
From observation 2, we observe,that reaction is zero order with respect to B.<o:p>
From observation 3, we observe that reaction is 2^nd order with respect to C.<o:p>Hence in total the reaction is of 3^rd order.

Alpha halogenation of carboxylic acid in presence of red phosphorus is called Hell Volhard Zelinsky reaction.

The correct relation between ∆H and ∆U is given by the relation
∆H = ∆U +∆n_gRT…………..where ∆n_g is the difference in number of moles of gaseous products and reactants. For the given reaction ∆ng = (2-[1+1]) = 0, so ∆H = ∆U.

For conversion of IO₄^- to I₂ in acidic medium the reaction is as under.<o:p>
IO₄^- + 7e^- → I₂……………… here oxidation state of I in IO₄^- is +7 and in I₂ is 0, so gain in number of electrons is of +7. We know that equilibrium weight is given by <o:p>
eqilibrium weight = molecular mass/gain in number of electrons<o:p> = M/7