AP EAMCET 2009 Chemistry Question Paper with Answer and Solution

(C) The formula for specific rotation is $[\alpha] = \frac{\alpha}{l \times c}$,where $\alpha$ is the observed rotation in degrees,$l$ is the path length in decimeters $(dm)$,and $c$ is the concentration in $g/mL$.
Given: $\alpha = -1.2^{\circ}$,$l = 5 \ cm = 0.5 \ dm$,and $c = 6.15 \ g / 100 \ mL = 0.0615 \ g/mL$.
Substituting these values: $[\alpha] = \frac{-1.2}{0.5 \times 0.0615} = \frac{-1.2}{0.03075} = -39^{\circ}$.

(A) To determine the structure of the alkene that undergoes ozonolysis,place the carbonyl oxygen atoms of the products face to face and replace the $O$ atoms with a double bond $(C=C)$.
Acetaldehyde is $CH_3CHO$ and acetone is $(CH_3)_2CO$.
Aligning them: $CH_3(H)C=O + O=C(CH_3)_2$.
Removing the oxygen atoms and joining the carbons with a double bond gives $CH_3(H)C=C(CH_3)_2$.
The structure is $CH_3-CH=C(CH_3)_2$.
The longest chain contains $4$ carbon atoms with a methyl group at the $2$-position,so the $IUPAC$ name is $2$-methyl-$2$-butene.

(C) Given,$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$.
Let $t = \tan \theta$,then $\theta = \tan^{-1} t$.
Then $\sqrt{1+t^2} = \sqrt{1+\tan^2 \theta} = \sec \theta$.
So,$x = \cos^{-1}\left(\frac{1}{\sec \theta}\right) = \cos^{-1}(\cos \theta) = \theta = \tan^{-1} t$.
And $y = \sin^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) = \sin^{-1}(\sin \theta) = \theta = \tan^{-1} t$.
Since $x = \tan^{-1} t$ and $y = \tan^{-1} t$,we have $y = x$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(A) Given,$z=\tan (y+a x)+\sqrt{y-a x}$
First,find the partial derivatives with respect to $x$:
$z_x = \sec^2(y+ax) \cdot a + \frac{1}{2\sqrt{y-ax}} \cdot (-a)$
$z_{xx} = \frac{\partial}{\partial x} [a \sec^2(y+ax) - \frac{a}{2\sqrt{y-ax}}] = 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}$
Next,find the partial derivatives with respect to $y$:
$z_y = \sec^2(y+ax) + \frac{1}{2\sqrt{y-ax}}$
$z_{yy} = \frac{\partial}{\partial y} [\sec^2(y+ax) + \frac{1}{2\sqrt{y-ax}}] = 2 \sec^2(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}$
Now,calculate $z_{xx} - a^2 z_{yy}$:
$z_{xx} - a^2 z_{yy} = [2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}] - a^2 [2 \sec^2(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}]$
$z_{xx} - a^2 z_{yy} = 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}} - 2a^2 \sec^2(y+ax) \tan(y+ax) + \frac{a^2}{4(y-ax)^{3/2}} = 0$

(B) Let the concentration of potassium acetate be $x \ M$. The mixture forms an acidic buffer. Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$
Given: $pH = 4.8$,$K_a = 1.8 \times 10^{-5}$,$pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
The number of moles of acid $= 20 \ mL \times 0.1 \ M = 2 \ mmol$.
The number of moles of salt $= 50 \ mL \times x \ M = 50x \ mmol$.
Substituting these into the equation:
$4.8 = 4.74 + \log \frac{50x}{2}$
$0.06 = \log(25x)$
$25x = 10^{0.06} \approx 1.148$
$x = \frac{1.148}{25} \approx 0.0459 \ M \approx 0.04 \ M$ (rounding to the nearest option).

(D) Buffer capacity,$\beta = \frac{dC_{HA}}{dpH}$.
Here,$dC_{HA}$ is the number of moles of acid added per liter of the solution.
$dC_{HA} = \frac{\text{moles of acetic acid}}{\text{volume in liters}} = \frac{0.12 / 60}{250 / 1000} = \frac{0.002}{0.25} = 0.008 \ M$.
The change in $pH$ is $dpH = 0.02$.
Therefore,$\beta = \frac{0.008}{0.02} = 0.4$.

(C) Given,$y=e^{a \sin ^{-1} x}$.
On differentiating with respect to $x$,we get:
$y_1 = e^{a \sin ^{-1} x} \cdot \frac{a}{\sqrt{1-x^2}}$
$\Rightarrow y_1 \sqrt{1-x^2} = ay$
Squaring both sides:
$(1-x^2) y_1^2 = a^2 y^2$
Again,differentiating with respect to $x$:
$(1-x^2) 2 y_1 y_2 - 2x y_1^2 = a^2 2y y_1$
Dividing by $2y_1$ (assuming $y_1 \neq 0$):
$(1-x^2) y_2 - x y_1 - a^2 y = 0$
Applying Leibnitz's theorem for the $n$-th derivative:
$(1-x^2) y_{n+2} + n(-2x) y_{n+1} + \frac{n(n-1)}{2}(-2) y_n - (x y_{n+1} + n(1) y_n) - a^2 y_n = 0$
$(1-x^2) y_{n+2} - 2nx y_{n+1} - n(n-1) y_n - x y_{n+1} - n y_n - a^2 y_n = 0$
$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} - (n^2 - n + n + a^2) y_n = 0$
$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} = (n^2 + a^2) y_n$

(B) Given,$f(x)=x^3+3x-2$.
On differentiating with respect to $x$,we get $f'(x)=3x^2+3$.
Since $x^2 \geq 0$ for all $x \in [2,3]$,$f'(x) = 3x^2+3 \geq 3 > 0$.
Thus,$f(x)$ is a strictly increasing function on the interval $[2,3]$.
For a strictly increasing function,the range is $[f(2), f(3)]$.
At $x=2$,$f(2) = 2^3 + 3(2) - 2 = 8 + 6 - 2 = 12$.
At $x=3$,$f(3) = 3^3 + 3(3) - 2 = 27 + 9 - 2 = 34$.
Therefore,the range of $f(x)$ is $[12, 34]$.

(D) Let $f(x) = \frac{\log x}{x}$.
To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x}{x^2} = \frac{1 - \log x}{x^2}$.
Set $f'(x) = 0$ to find critical points:
$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
Now,we check the second derivative $f''(x)$ to confirm the nature of the point:
$f''(x) = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.
At $x = e$,$f''(e) = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3} < 0$.
Since the second derivative is negative,the function has a local maximum at $x = e$.
The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e} = e^{-1}$.

(A) We have the integral $I = \int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$.
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 - \cos 2x = 2 \sin^2 x$,we get:
$I = \int \left( \frac{2 - 2 \sin x \cos x}{2 \sin^2 x} \right) e^x \, dx$
$I = \int \left( \frac{1}{\sin^2 x} - \frac{\sin x \cos x}{\sin^2 x} \right) e^x \, dx$
$I = \int (\operatorname{cosec}^2 x - \cot x) e^x \, dx$
$I = \int e^x \operatorname{cosec}^2 x \, dx - \int e^x \cot x \, dx$
Using integration by parts for the first term $\int e^x \operatorname{cosec}^2 x \, dx$:
Let $u = \cot x$,then $du = -\operatorname{cosec}^2 x \, dx$.
$\int e^x \operatorname{cosec}^2 x \, dx = -e^x \cot x - \int (-e^x \cot x) \, dx = -e^x \cot x + \int e^x \cot x \, dx$.
Substituting this back into the expression for $I$:
$I = (-e^x \cot x + \int e^x \cot x \, dx) - \int e^x \cot x \, dx + c$
$I = -e^x \cot x + c$.

(D) Given the family of curves: $y = (a + bx + cx^2)e^x$
This can be written as $y e^{-x} = a + bx + cx^2$.
Differentiating with respect to $x$ three times:
First derivative: $\frac{d}{dx}(y e^{-x}) = b + 2cx$
$\Rightarrow y' e^{-x} - y e^{-x} = b + 2cx$
Second derivative: $\frac{d}{dx}(y' e^{-x} - y e^{-x}) = 2c$
$\Rightarrow y'' e^{-x} - y' e^{-x} - (y' e^{-x} - y e^{-x}) = 2c$
$\Rightarrow y'' e^{-x} - 2y' e^{-x} + y e^{-x} = 2c$
Third derivative: $\frac{d}{dx}(y'' e^{-x} - 2y' e^{-x} + y e^{-x}) = 0$
$\Rightarrow y''' e^{-x} - y'' e^{-x} - 2(y'' e^{-x} - y' e^{-x}) + (y' e^{-x} - y e^{-x}) = 0$
$\Rightarrow y''' - y'' - 2y'' + 2y' + y' - y = 0$
$\Rightarrow y''' - 3y'' + 3y' - y = 0$
Thus,the correct option is $D$.

(B) Given the differential equation: $\frac{dy}{dx} = \sin(x+y) \tan(x+y) - 1$
Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$,which implies $\frac{dy}{dx} = \frac{dz}{dx} - 1$.
Substituting this into the original equation:
$\frac{dz}{dx} - 1 = \sin z \tan z - 1$
$\frac{dz}{dx} = \sin z \tan z = \sin z \cdot \frac{\sin z}{\cos z} = \frac{\sin^2 z}{\cos z}$
Rearranging the terms to separate variables:
$\int \frac{\cos z}{\sin^2 z} dz = \int dx$
Let $\sin z = t$,then $\cos z dz = dt$. The integral becomes:
$\int \frac{1}{t^2} dt = x + c$
$- \frac{1}{t} = x + c$
Since $t = \sin z$,we have $-\operatorname{cosec} z = x + c$,which simplifies to $x + \operatorname{cosec} z = -c$. Replacing $c$ with a new constant $C$,we get:
$x + \operatorname{cosec}(x+y) = C$

(C) We know that for a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines are $\cos \alpha, \cos \beta, \cos \gamma$,and they satisfy the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given expression: $E = \cos 2\alpha + \cos 2\beta + \cos 2\gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we can rewrite the expression as:
$E = (\cos^2 \alpha - \sin^2 \alpha) + (\cos^2 \beta - \sin^2 \beta) + (\cos^2 \gamma - \sin^2 \gamma) + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Simplifying the terms:
$E = \cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta + \cos^2 \gamma - \sin^2 \gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
The terms $-\sin^2 \alpha + \sin^2 \alpha$,$-\sin^2 \beta + \sin^2 \beta$,and $-\sin^2 \gamma + \sin^2 \gamma$ cancel out.
Thus,$E = \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$.
Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,the value of the expression is $1$.

(C) Given: $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$,$P(E_1 / E_2) = \frac{1}{4}$.
Step $1$: Find $P(E_1 \cap E_2)$.
$P(E_2 / E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)} \Rightarrow \frac{1}{2} = \frac{P(E_1 \cap E_2)}{1/4} \Rightarrow P(E_1 \cap E_2) = \frac{1}{8}$.
Step $2$: Find $P(E_2)$.
$P(E_1 / E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} \Rightarrow \frac{1}{4} = \frac{1/8}{P(E_2)} \Rightarrow P(E_2) = \frac{1/8}{1/4} = \frac{1}{2}$. (Matches $iv$)
Step $3$: Find $P(E_1 \cup E_2)$.
$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$. (Matches $ii$)
Step $4$: Find $P(\bar{E}_1 / \bar{E}_2)$.
$P(\bar{E}_1 / \bar{E}_2) = \frac{P(\bar{E}_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{1 - P(E_1 \cup E_2)}{1 - P(E_2)} = \frac{1 - 5/8}{1 - 1/2} = \frac{3/8}{1/2} = \frac{3}{4}$. (Matches $vi$)
Step $5$: Find $P(E_1 / \bar{E}_2)$.
$P(E_1 / \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{1 - P(E_2)} = \frac{1/4 - 1/8}{1 - 1/2} = \frac{1/8}{1/2} = \frac{1}{4}$. (Matches $i$)
Thus,the correct matching is $(A)$-$iv$,$(B)$-$ii$,$(C)$-$vi$,$(D)$-$i$.

(A) Given that $X$ is a binomial variate with range $\{0, 1, 2, 3, 4, 5, 6\}$,so $n = 6$.
The probability mass function of a binomial distribution is given by $P(X=k) = { }^n C_k p^k q^{n-k}$,where $q = 1-p$.
According to the given condition,$P(X=2) = 4 P(X=4)$.
Substituting the values:
${ }^6 C_2 p^2 q^4 = 4 \cdot { }^6 C_4 p^4 q^2$
Since ${ }^6 C_2 = 15$ and ${ }^6 C_4 = 15$,we have:
$15 p^2 q^4 = 4 \cdot 15 p^4 q^2$
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$q^2 = 4 p^2$
$(1-p)^2 = 4 p^2$
$1 - 2p + p^2 = 4 p^2$
$3p^2 + 2p - 1 = 0$
Factoring the quadratic equation:
$(3p - 1)(p + 1) = 0$
This gives $p = \frac{1}{3}$ or $p = -1$.
Since the probability $p$ must be in the interval $[0, 1]$,we discard $p = -1$.
Therefore,$p = \frac{1}{3}$.

(C) Aluminium reacts with aqueous $NaOH$ to form sodium tetrahydroxoaluminate$(III)$.
The reaction is: $2Al(s) + 2NaOH(aq) + 6H_2O(l) \longrightarrow 2Na[Al(OH)_4](aq) + 3H_2(g)$.
In aqueous solution,the species exists as the octahedral complex $[Al(H_2O)_2(OH)_4]^-$,where the coordination number of $Al$ is $6$.

(D) The hydrolysis of silicon tetrachloride $(SiCl_4)$ proceeds as follows:
$SiCl_4 + 4H_2O \longrightarrow H_4SiO_4 + 4HCl$
Here,'$X$' is silicic acid $(H_4SiO_4)$.
Upon heating at $1000^{\circ} C$,silicic acid undergoes dehydration:
$H_4SiO_4 \xrightarrow{\Delta, 1000^{\circ} C} SiO_2 + 2H_2O$
Here,'$Y$' is silicon dioxide $(SiO_2)$.
Therefore,'$X$' is $H_4SiO_4$ and '$Y$' is $SiO_2$.

(C) Assertion $(A)$ is true: Alkali metals $K$,$Rb$,and $Cs$ react with excess oxygen to form superoxides of the type $MO_2$.
Reason $(R)$ is false: The stability of superoxides increases from $K$ to $Cs$ because the large size of the alkali metal cation stabilizes the large superoxide anion $(O_2^-)$ through a decrease in lattice energy,but the statement claims stability increases *due to* the decrease in lattice energy,which is a misinterpretation of the thermodynamic stability factor. More accurately,the stability increases due to the increasing size of the cation which matches the size of the large anion,reducing the lattice energy and making the formation more favorable. However,the reason provided is scientifically incorrect in its causal link.
Therefore,$(A)$ is true but $(R)$ is false.

(C) Mass of $Cd = 0.9 \ g$.
Mass of $Cl_2 = 1.5 \ g - 0.9 \ g = 0.6 \ g$.
Atomic weight of $Cl = 35.5 \ g/mol$.
Mass of $Cl_2$ in $CdCl_2 = 2 \times 35.5 = 71 \ g/mol$.
According to the law of chemical equivalence:
$\frac{\text{Mass of } Cd}{\text{Atomic weight of } Cd} = \frac{\text{Mass of } Cl_2}{\text{Equivalent weight of } Cl_2}$.
$\frac{0.9}{x} = \frac{0.6}{71}$.
$x = \frac{0.9 \times 71}{0.6} = 1.5 \times 71 = 106.5 \ g/mol$.

(B) Step $1$: Determine the empirical formula of $A$.
$C = 85.71 \% = \frac{85.71}{12} = 7.14$; $\frac{7.14}{7.14} = 1$
$H = 14.29 \% = \frac{14.29}{1} = 14.29$; $\frac{14.29}{7.14} = 2$
Empirical formula $= CH_2$.
Step $2$: Determine the molecular formula of $A$.
Molecular weight $= 2 \times \text{vapour density} = 2 \times 14 = 28$.
$n = \frac{28}{14} = 2$.
Molecular formula $= (CH_2)_2 = C_2H_4$ (Ethene).
Step $3$: Reaction sequence.
$A$ is $CH_2=CH_2$.
$CH_2=CH_2 + Cl_2/H_2O \rightarrow HO-CH_2-CH_2-Cl$ ($B$,Ethylene chlorohydrin).
$HO-CH_2-CH_2-Cl + KCN \rightarrow HO-CH_2-CH_2-CN$ (Nucleophilic substitution).
$HO-CH_2-CH_2-CN + H_3O^+ \rightarrow HO-CH_2-CH_2-COOH$ ($C$,$3$-hydroxypropanoic acid).

(A) Perhydrol is a $30\% \ w/v$ $H_2O_2$ solution,which corresponds to $100$ volume $H_2O_2$.
This means $1 \ mL$ of perhydrol produces $100 \ mL$ of $O_2$ at $STP$.
The reaction for conversion is: $2SO_2 + O_2 \rightarrow 2SO_3$.
From the stoichiometry,$2 \ L$ of $SO_2$ requires $1 \ L$ of $O_2$ for complete conversion.
Volume of $O_2$ required = $1 \ L = 1000 \ mL$.
Therefore,volume of perhydrol required = $\frac{1000 \ mL}{100} = 10 \ mL$.

(B) The average kinetic energy per molecule of an ideal gas is given by the formula: $KE_{avg} = \frac{3}{2} k T$.
Here,$k$ is the Boltzmann constant $(k = \frac{R}{N_A})$,$T$ is the temperature in Kelvin,and $N_A$ is Avogadro's number.
Given: $T = 27^{\circ} C = 27 + 273 = 300 \ K$.
Substituting the values:
$KE_{avg} = \frac{3}{2} \times \left( \frac{8.314 \ J \ K^{-1} \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} \right) \times 300 \ K$.
$KE_{avg} = 1.5 \times 1.38 \times 10^{-23} \ J \ K^{-1} \times 300 \ K$.
$KE_{avg} \approx 6.21 \times 10^{-21} \ J \ \text{molecule}^{-1}$.

(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 \times 10^4 \ J$.
Since $1 \ mol = 6.023 \times 10^{23} \ \text{atoms}$,the $KE$ of $1 \ \text{electron}$ is:
$KE = \frac{6.023 \times 10^4 \ J}{6.023 \times 10^{23}} = 1.0 \times 10^{-19} \ J$.
The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$:
$E = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{600 \times 10^{-9} \ m} = 3.313 \times 10^{-19} \ J$.
The threshold energy $(Phi)$ is the difference between the incident photon energy and the kinetic energy:
$\Phi = E - KE = 3.313 \times 10^{-19} \ J - 1.0 \times 10^{-19} \ J = 2.313 \times 10^{-19} \ J$.

(A) From de-Broglie's equation,$\lambda = \frac{h}{mv}$.
Squaring both sides,$\lambda^2 = \frac{h^2}{m^2v^2}$.
Rearranging for kinetic energy $(KE = \frac{1}{2}mv^2)$,we get $mv^2 = \frac{h^2}{m\lambda^2}$.
Therefore,$KE = \frac{1}{2} \times \frac{h^2}{m\lambda^2}$.
This implies $KE \propto \frac{1}{\lambda^2}$.
Given the ratio of wavelengths $\frac{\lambda_1}{\lambda_2} = \frac{3}{5}$,the ratio of kinetic energies is $\frac{K_1}{K_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$.
Thus,the ratio is $25: 9$.

(C) For cylinder $A$ (isobaric process): Heat given $Q = n C_P \Delta T_A$. Given $\Delta T_A = 42 ~K$ and for monoatomic gas $C_P = \frac{5}{2} R$. Thus,$Q = n \left(\frac{5}{2} R\right) (42) = 105 nR$.
For cylinder $B$ (isochoric process): Heat given $Q = n C_V \Delta T_B$. For monoatomic gas $C_V = \frac{3}{2} R$.
Since the heat given is the same,$n \left(\frac{3}{2} R\right) \Delta T_B = 105 nR$.
$\frac{3}{2} \Delta T_B = 105 \implies \Delta T_B = 105 \times \frac{2}{3} = 70 ~K$.

(B) To obtain the target reaction $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$,we manipulate the given equations:
$1$. Multiply equation $(A)$ by $4$: $4Na_{(s)} + 4H_2O_{(l)} \longrightarrow 4NaOH_{(s)} + 2H_{2(g)} \quad \Delta H_1 = 4 \times (-146) = -584 \ kJ$
$2$. Reverse equation $(B)$: $2NaOH_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)} + H_2O_{(l)} \quad \Delta H_2 = -418 \ kJ$
$3$. Keep equation $(C)$ as is: $2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H_3 = +259 \ kJ$
Adding these three equations:
$(4Na + 4H_2O) + (2NaOH + SO_3) + (2Na_2O + 2H_2) \longrightarrow (4NaOH + 2H_2) + (Na_2SO_4 + H_2O) + (4Na + 2H_2O)$
Simplifying the equation gives: $2Na_2O_{(s)} + 2SO_{3(g)} \longrightarrow 2Na_2SO_{4(s)}$
Dividing by $2$: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$
The enthalpy change is $\Delta H = \frac{1}{2} (\Delta H_1 + \Delta H_2 + \Delta H_3) = \frac{1}{2} (-584 - 418 + 259) = \frac{1}{2} (-743) = -371.5 \ kJ$.
Wait,re-evaluating the combination:
$2 \times (A) + \frac{1}{2} \times (C) - (B) = 2(-146) + \frac{1}{2}(259) - 418 = -292 + 129.5 - 418 = -580.5 \ kJ \approx -581 \ kJ$.

(C) Given: $\Delta H_f(H) = 218 \ kJ/mol$
The reaction for the formation of one mole of $H$ atoms is: $\frac{1}{2} H_2 \rightarrow H ; \Delta H = 218 \ kJ/mol$
The bond dissociation energy of $H-H$ is the energy required for the reaction: $H_2 \rightarrow 2H$
Therefore,$\Delta H_{bond} = 2 \times 218 \ kJ/mol = 436 \ kJ/mol$
To convert $kJ/mol$ to $kcal/mol$,we use the conversion factor $1 \ kcal = 4.18 \ kJ$
$\Delta H_{bond} = \frac{436}{4.18} \ kcal/mol \approx 104.3 \ kcal/mol$
Thus,the $H-H$ bond energy is approximately $104 \ kcal/mol$.

(A) The position of the $n^{\text{th}}$ bright fringe (maximum) from the central maximum in Young's double slit experiment is given by the formula:
$y_n = \frac{n \lambda D}{d}$
where $D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
For the first case,the $10^{\text{th}}$ maximum of wavelength $\lambda_1$ is at distance $y_1$:
$y_1 = \frac{10 \lambda_1 D}{d}$
For the second case,the $5^{\text{th}}$ maximum of wavelength $\lambda_2$ is at distance $y_2$:
$y_2 = \frac{5 \lambda_2 D}{d}$
Now,calculating the ratio $\left(\frac{y_1}{y_2}\right)$:
$\frac{y_1}{y_2} = \frac{\frac{10 \lambda_1 D}{d}}{\frac{5 \lambda_2 D}{d}}$
$\frac{y_1}{y_2} = \frac{10 \lambda_1}{5 \lambda_2} = \frac{2 \lambda_1}{\lambda_2}$

(A) Let the intensity at the central bright fringe be $I_1 = I_{max} = 4I_0$,where $I_0$ is the intensity of each individual slit.
The path difference $\Delta x$ at a distance $y$ from the central maximum is given by $\Delta x = \frac{yd}{D}$.
The fringe width is $\beta = \frac{\lambda D}{d}$.
Given that $y = \frac{\beta}{4} = \frac{\lambda D}{4d}$,the path difference at $P_2$ is $\Delta x = \frac{(\lambda D / 4d)d}{D} = \frac{\lambda}{4}$.
The corresponding phase difference is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.
The intensity at any point is given by $I = I_{max} \cos^2(\phi / 2)$.
Substituting $\phi = \frac{\pi}{2}$,we get $I_2 = I_{max} \cos^2(\frac{\pi}{4}) = I_{max} (\frac{1}{\sqrt{2}})^2 = \frac{I_{max}}{2}$.
Therefore,$\frac{I_1}{I_2} = \frac{I_{max}}{I_{max}/2} = 2$.