Class 12th

26.  (ii) Free valence electrons 

4. Iron oxides are mixed with limestone and coke in order to decompose carbonates and oxidize sulfides. This mixture is then fed into the blast furnace. Many reactions take place in different temperature ranges of the blast furnace. By oxidizing impurities from cast iron in a reverberatory furnace which is lined by hematite, wrought iron or malleable iron is obtained. This is the purest form of commercial iron. This can be shown with the following reaction:

Fe2O₃ + 3C → 2Fe + 3CO.

Sulfur, silicone and phosphorus are oxidized and passed into the slag and removed as impurities when limestone is added as a flux to it.

3. From Ellingham diagram, we know that ΔG^° _{(C, CO)}  < G^° _{(Fe, Fe)}, the following reactions are:

C+ $\frac{1}{2}$ O₂→CO

2Fe+O₂→2FeO

Therefore, FeO can be reduced to Fe by coke. 

25. Option (i) Very low value of electrical conductivity in the molten state is correct since in molten state, the ionic solids possess free ions which help in the conduction of electricity. Due to the availability of these free ions the electrical conductivity of ionic compounds in molten state is found to be very high. 

2. Extraction of chlorine from brine is an oxidation method. The reactions involved are:

2Cl⁻ + 2H₂O → 2OH⁻ + H₂ + Cl₂

For this reaction, the value of ΔG°=+422 kJ, which is positive. Using the formula ΔG°=−nE°F, we get a negative value of E° =−2.2 V.

Since,  E°=−2.2 V, it would naturally require an external voltage greater than 2.2 V for the reaction to occur.

24. Option (iii) (C) and (D) is correct since Iodine is a non-polar molecular solid which acts as an insulator while H₂O is a hydrogen bonded molecular solid which also acts as an insulator. Whereas Mg is a metal and acts as a conductor in solid and in its molten state and TiO is also an ionic solid which acts as an insulator in solid state but acts as a conductor in molten and in aqueous solutions. 

The given differential equation is:

$\begin{array}{l}{e}^{x}dy+(y{e}^x+2x)dx=0\\ \Rightarrow e^x\frac{dy}{dx}+y{e}^x+2x=0\\ \Rightarrow \frac{dy}{dx}+y=-2x{e}^{-x}\end{array}$

This is a linear differential equation of the form

$\begin{array}{l}\frac{dy}{dx}+Py=Q,whereP=1\&Q=-2x{e}^{-x}\\ Now,I.F.=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int dx}}=e^x\end{array}$

The general solution of the given differential equation is given by,

$\begin{array}{l}y(I.F.)={\displaystyle \int \left(Q*I.F.\right)dx+C}\\ \Rightarrow y{e}^x={\displaystyle \int (-2x{e}^{-x}.e^x)dx+C}\\ \Rightarrow y{e}^x=-{\displaystyle \int 2xdx+C}\\ \Rightarrow y{e}^x=-x^2+C\\ \Rightarrow y{e}^x+x^2=C\end{array}$

Therefore, option (c) is correct.

The integrating factor of the given differential equation  $\frac{dx}{dy}+P_{1}x=Q_1$

The general solution of the differential equation is given by,

$\begin{array}{l}x(I.F.)={\displaystyle \int \left(Q*I.F.\right)dy+C}\\ \Rightarrow x.e^{{\displaystyle \int P_{1}dy}}={\displaystyle \int \left(Q_1{e}^{{\displaystyle \int P_{1}dy}}\right)}dy+C\end{array}$

Hence, the correct answer is C.

The given differential equation is:

$\begin{array}{l}\frac{ydx-xdy}{y}=0\\ \Rightarrow \frac{ydx-xdy}{xy}=0\\ \Rightarrow \frac{1}{x}dx-\frac{1}{y}dy=0\end{array}$

Integration both sides, we get:

$\begin{array}{l}log\left|x\right|-log\left|y\right|=logk\\ \Rightarrow log\left|\frac{x}{y}\right|=logk\\ \Rightarrow \frac{x}{y}=k\\ \Rightarrow y=\frac{1}{k}x\\ \Rightarrow y=Cx,where,C=\frac{1}{k}\end{array}$

Therefore, option (C) is correct.

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$\therefore \frac{dy}{dt}\alpha y$

$\Rightarrow \frac{dy}{dt}=ky$ (k is constant)

$\Rightarrow \frac{dy}{y}=kdt$

Integration both sides, we get:

$logy=kt+C..........(1)$

In the year $1999,t=0\&y=20000.$

Therefore, we get:

$log20000=C..........(2)$

In the year $2004,t=5\&y=25000.$

Therefore, we get:

$\begin{array}{l}\Rightarrow log25000=5k+log20000\\ \Rightarrow 5k=log\left(\frac{25000}{20000}\right)=log\left(\frac{5}{4}\right)\\ \Rightarrow k=\frac{1}{5}log\left(\frac{5}{4}\right)..........(3)\end{array}$

In the year $2009,t=10years$

Now, on substituting the values of t, k, and C in equation (1), we get:

$\begin{array}{l}logy=10*\frac{1}{5}log\left(\frac{5}{4}\right)+log(20000)\\ \Rightarrow logy=log\left[20000*{\left(\frac{5}{4}\right)}^2\right]\\ \Rightarrow y=20000*\frac{5}{4}*\frac{5}{4}\\ \Rightarrow y=31250\end{array}$

Hence, the population of the village in 2009 will be 31250.