Chemistry Test - 40

Acid containing Oxygen is Oxo-Acids.

An oxyacid, oxoacid, or ternary acid is an acid that contains oxygen. Specifically, it is a compound that contains hydrogen, oxygen, and at least one other element, with at least one hydrogen atom bond to oxygen that can dissociate to produce the H⁺ cation and the anion of the acid. e.g., carbonic acid (H₂CO₃), sulphuric acid (H₂SO₄).

In \(A g_{2} S O_{4}\), silver is present as \(A g^{+}\), in \(C u F_{2}\), copper is present as \(C u^{2+}\), in \(Z n F_{2}\), zinc is present as \(Z n^{2+}\), in \(C u_{2} C l_{2}\), the copper is present as \(C u^{+1}\). 

As they are metal cations, the electrons are removed from their actual configuration to form ions.

The respective transition elements exists in:

\(A g^{+}-[A r]4 d^{10} 5 s^{0},\) (have completely filled \(d\) orbital).

\(C u^{+2}-[A r]3 d^{9} 4 s^{0},\) (have incompletely filled \(d\) orbital).

\({Zn}^{+2}-[A r]3 d^{10} 4 s^{0},\) (have completely filled \(d\) orbital).

\({Cu}^{+1}-[A r]3 d^{10} 4 s^{9},\) (have completely filled \({d}\) orbital). 

Since, \(C u^{+2}\) has unpaired electron, it will show electron transitions. Therefore, will be coloured.

In order to calculate the enthalpy change for \(\mathrm{H}_2 \mathrm{O}\) at \(5^{\circ} \mathrm{C}\) to ice at \(-5^{\circ} \mathrm{C}\), we need to calculate the enthalpy change of all the transformation involved in the process.

(a) Energy change of \(1 \mathrm{~mol}, \mathrm{H}_2 \mathrm{O}(1)\), at \(5^{\circ} \mathrm{C}\)

\(\rightarrow 1 \mathrm{~mol}, \mathrm{H}_2 \mathrm{O}(\mathrm{I}), 0^{\circ} \mathrm{C}\)

(b) Energy change of \(1 \mathrm{~mol}, \mathrm{H}_2 \mathrm{O}(1)\), at \(0^{\circ} \mathrm{C}\)

\(\rightarrow 1 \mathrm{~mol}, \mathrm{H}_2 \mathrm{O} \text { (s)(ice), } 0^{\circ} \mathrm{C}\)

(c) Energy change of \(1 \mathrm{~mol}\), ice (s), at \(0^{\circ} \mathrm{C}\)

\(\rightarrow 1 \mathrm{~mol} \text {, ice }(\mathrm{s})_2-5^{\circ} \mathrm{C}\)

\(\text { Total } \Delta \mathrm{H}\)

\(=\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(1)\right] \Delta \mathrm{T}+\Delta \mathrm{H} \text { freezing }+\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{s})\right] \Delta \mathrm{T} \)

\(=\left(75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(0-5) \mathrm{K}+\left(-6 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right) \)

\(+\left(36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(-5-0) \mathrm{K} \)

\(\Delta \mathrm{H}=-6.56 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { (cxothermic process) }\)

\(4\)isomers are possible for \(\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{~N}\).

Given,

Frequency, \(\nu=5 \times 10^{14}\) \(\mathrm{Hz}\)

As we know, Planck constant is given as,

\(h=6.626 \times 10^{-34} \mathrm{Js}\)

Energy of one photon is given by theexpression,

\(E=h \nu\)

\(E=\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(5 \times 10^{14} \mathrm{~s}^{-1}\right)\)

\(=3.313 \times 10^{-19} \mathrm{~J}\)

As we know,

Avogadro's number \(= 6.022 × 10^{23} \mathrm{~mol}^{-1}\)

Energy of one mole of photons,

\(=\left(3.313 \times 10^{-19} \mathrm{~J}\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)\)

\(=199.51 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

A system is said to be in thermodynamic equilibrium if the conditions for the following all three equilibrium is satisfied:

Mechanical equilibrium:

When there are no unbalanced forces within the system and between the system and the surrounding, the system is said to be under mechanical equilibrium.

Chemical equilibrium:

The system is said to be in chemical equilibrium when there are no chemical reactions going on within the system or there is no transfer of matter from one part of the system to another due to diffusion.

Thermal equilibrium:

When the temperature of the system is uniform and not changing throughout the system and also in the surroundings, the system is said to be in thermal equilibrium.

Hydrogen gas is not obtained when zinc reacts with cold water.

Zinc is a low reactive metal which does not react with cold water and hence no gas is evolved. However, \(\mathrm{Zn}\) reacts with dil. \(\mathrm{HCl}\), dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and hot \(\mathrm{NaOH}\) solution.

Reaction (a) gives

$2H_{2}O\left(l\right)\underset{Traces of acid /base}{\overset{Electrolysis}{\to }}2H_{2}O\left(g\right)+O_2\left(g\right)$

Reaction (b) gives

$By electrolysis \to of brine solution$

At anode: $2C{l}^{-}\left(aq\right) \to C{l}_2\left(g\right)+2e^{-}$

At Cathode: $2H_{2}O\left(l\right)+2e^{-}\to H_{2}O\left(g^{}\right)+2O{H}^{-}\left(aq\right)$

Overall reaction:

Reaction (c) gives

$C_n{H}_{2n+2}+ n{H}_{2}O \underset{Ni}{\overset{1270 K}{\to }}nCO +\left(3n+1\right)H_2$

The mixture of CO and water is termed as syn gas.

If \(a\) is the edge length of the cube and \(r\) is the radius of each atom.

Distance of nearest approach between two atoms \(=2 r=1.73 \)Å.

Therefore,

\(r=\frac{1.73}{2}=0.865 \)Å

For BCC lattice, the relation of edge length and radius of the atom is,

\(\sqrt{3} a=4 r\)

\(a=\frac{4 r}{\sqrt{3}}\)

\(a=\frac{4 \times 0.865}{\sqrt{3}}\)

\(a=1.99=2\)Å \(=200 \mathrm{pm}\)

The edge length of the BCC lattice is \(200 \mathrm{pm}\).

In the structure of \(C r O_{5}\), it has four \(O\) atoms as peroxide which have oxidation number \(=-1\) and one \(O\) as oxide (double bonded \(O)\) atom with oxidation number \(=-2\).

Let the oxidation number of \(C r\) is \(x\).

Then, \(x+4 \times(-1)+1 \times(-2)=0\) or, \(x=+6\)

The oxidation number of \(C r\) in \(C r O_{5}\) is \(+6\).