Chemistry Test-2

Since, alkene shows geometrical isomerism. So, the groups should be different on both sides of double bond.

When 2-butanol is heated with conc. sulfuric acid, it forms 2-butene, which shows geometrical isomerism in the form of cis and trans isomers.

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} \underset{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\stackrel{\text { dehydration }}{\longrightarrow}} \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\)

H₂S is less acidic than H₂Te because as we move down the group, the bond dissociation enthalpy decreases i.e it becomes easy to dissociate or break the bonds.

Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat (is negative) and an increase in entropy (is positive).

Given: Vapour pressure of liquid A, \(P_{A} =120~ mm\)

Vapour pressure of liquid B, \(P_{B} =180~ mm\)

Number of moles of liquid A,  \({n}_{{A}}=2\)

Number of moles of liquid B, \({n}_{{B}}=3\)

Therefore,

Mole fraction of liquid A,  \({X}_{{A}}=\frac{n_A}{n_A+n_B}\)\(=\frac{2}{5}\)

Mole fraction of liquid B,  \({X}_{{B}}=\frac{n_B}{n_A+n_B}\)\(= \frac{3}{5}\)

We know that: Vapor Pressure of solution, \({P}={X}_{{A}} {P}_{{A}}+{X}_{{B}} {P}_{{B}}\)

\(=\frac{2}{5} \times 120+\frac{3}{5} \times 180\)\(=48+108\)

\(P=156 {~mm} {~Hg}\)

Given,

Temperature \(T =45^{\circ} C\)

We know, \(n\) (number of electrons transferred) \(=2\)

According to Nernst equation, \(E _{(\frac{ M^{ 2+}} {M} )}= E _{( \frac{M^{ 2+}}{ M} )}+2.303 \frac{R T}{n F} \log _{10}\left( \frac{M} {[ M ]^{+2}}\right)\)

Concentration of \([ M ]\) is taken to be \(1\)

The equation becomes: \(E ^{\circ}( \frac{M^{ 2+}}{M} )+2.303 \frac{R T}{n F} \log _{10}\left(\frac{1} {[ M ]^{+2}}\right)\)

\(E _{(\frac{ M^{ 2+}}{ M} )}= E _{(\frac{ M^{ 2+}}{ M} )}+2.303 \times \frac{8.314 \times 318}{2 \times 96500} \log _{10}\left(\frac{1} {[ M ]^{+2}}\right)= E _{( \frac{M^{ 2+}}{ M} )}+0.0315 \log _{10}\left(\frac{1}{[ M ]^{+2}}\right)\)

The phenomenon that takes place at the surface, is termed a surface phenomenon. Among the given processes, crystallization takes place in the bulk, not at the surface, so it is not a surface phenomenon.

The most electronegative element from the following elements is fluorine.

Fluorine is the most electronegative element. It has an atomic configuration of \(1 s^2 2 s^2 2 p^5\). The \(s\) subshell has \(1\) orbital that can hold up to \(2\) electrons, the \(p\) subshell has \(3\) orbitals that can hold up to \(6\) electrons, the \(d\) subshell has \(5\) orbitals that hold up to \(10\) electrons, and the \(f\) subshell has \(7\) orbitals with \(14\) electrons. There are \(5\) electrons in \(p\) subshell which means it needs only one \(1\) electron to complete the orbit so it will try to accept an electron which makes flourine the most electronegative element

The electronegativity is the property of an atom that increases with its tendency to attract the electrons of a bond.Moving from left to right in any period, the electronegativity increases and moving from top to bottom in any group, the electronegativity decreases.

The primary and secondary valency of cobalt ion is 3 and 6 respectively.

The charge on cobalt ion is +3 which balances -2 charge on carbonate ion and -1 charge on chloride ion. Therefore, the primary valency of cobalt ion is 3.

Cobalt ion is surrounded by four monodentate and one bidentate ligand (4 ammonia and on carbonate) in the coordination sphere. Therefore, its coordination number (secondary valency) is 6.

On hydrolysis of starch, we finally get glucose. This is because starch is the polymer of glucose. The reaction takes place in presence of amylase enzymes.

Synthetic detergents are cleansing agents which have all the properties of soaps, but which actually do not contain any soap. These can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water. Synthetic detergents are mainly classified into three categories:

(i) Anionic detergents

(ii) Cationic detergents and

(iii) Non-ionicdetergents

Given,

The brown ring complex compound of iron is formulated as \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{NO})\right] \mathrm{SO}_{4}\).

Applying formula,

Sum of total oxidation state of all atoms = Overall charge on the compound

Let oxidation state of iron is \(\mathrm{x}\).

\(\mathrm{x}+0(5)+1(1)+1(-2)=0\)

\(\therefore \mathrm{x}=+1\)

Fe is in \(+1\) oxidation state. \(\mathrm{H}_{2} \mathrm{O}\) is a neutral ligand. \(\mathrm{NO}^{+}\) is a positive charge ligand.