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Intermolecular H- bonding and intra-molecular H- bonding producing compound may be the phenol derivatives.

E₁ (done to q₁) = $\frac{k,q_1}{{\left(d/2\right)}^2}=\frac{\left[9*10^9*8*10^{?6}\right]}{d^2}*4$

E₂ (done to q₂) = $\frac{k\text{?}{q}_2}{{\left(d/2\right)}^2}=\frac{\left[9*10^9*8*10^{?6}\right]}{d^2}*4$

Enet = E₁ + E₂

$=2\left[\frac{4*9*10^9*8*10^{?6}}{d^2}\right]$

Given Enet = 6.4 * 104

$?2*\left[\frac{4*9*10^9*8*10^{?6}}{d^2}\right]=6.4*10^4$

d² = 9 * 10^-6 + 6

d = 3m

Complete combustion of compound produces 0.2 gm CO₂

Hence wt of carbon in 0.2 gm CO₂

$=\left(\frac{12}{44}\times 0.2\right)gm$

Therefore % of carbon in compound

$=\frac{wtofcarbon\times 100}{wtofcompound}=\frac{12\times 0.2\times 100}{44\times 0.3}=\frac{2400}{44\times 3}=18.18\approx 18\mathrm{\%}$

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In the VTUEEE the course are only for B.Tech course, and for this course only three subjects are mandatory - Physics, Chemistry and Mathematics. Hnece there is no scope for students to study Biology subject.

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For precipitation of two moles of AgCl

Two Cl will produce as a free anion

CoCl₃.4NH₃ complex will $\left[CO{\left(N{H}_3\right)}_{4}C{l}_2\right]$ Cl (will not give 2Cl)

$PtC{l}_4.2HCl\to$ complex will be H2 [PtCl6] will not any Cl

$NiC{l}_2.6H_{2}O\to \left[Ni{\left(H_{2}O\right)}_6\right]C{l}_2$ will produce two Cl ion.

${\left[Ni{\left(H_{2}O\right)}_6\right]}^{++}+2C{l}^{-}\underset{AgN{O}_3}{\overset{}{\to }}2AgCl\left(s\right)$ precipitate formation

$N{i}^{2+}\to \left[Ar\right]3d^{8}4s^0$

$\mu =\sqrt[]{2\left(2+2\right)}=\sqrt[]{8}=2.84BM=3$

V_rms = $\sqrt[]{\frac{3RT}{M}}$

V_rms’ = $\sqrt[]{\frac{3R{T}_f}{m_f}}=\sqrt[]{\frac{3R\times 2T\times 2}{M}}$ $\left[\begin{array}{cc}\hfill Given,T_f=2T\hfill & \hfill M_f=\frac{M}{2}\hfill \end{array}\right]$

$=2\sqrt[]{\frac{3RT}{M}}$

V_rms’ = 2V_rms

Most basic oxide V₂O3

Here V has +3 O.S. Hence V⁺³ $\left[Ar\right]3d^2$

two unpaired e^- in d- subshell

$\mu =\sqrt[]{2\left(2+2\right)}=\sqrt[]{8}=2.84BM\approx 3$

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Volume of H₂ adsorbed = $\frac{nRT}{P}=\frac{2\times 0.083\times 300}{2\times 1}=24.9lit=24900ml$

Therefore volume of gas adsorbed per gram of the adsorbent = $\frac{24900}{2.5}=9960$

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BTech is a four year undergraduate programme that is designed to focus on technical and engineering education. BTech curriculum usually includes core engineering subjects like Mathematics and Programming. Students also go industrial training and internships for marketplace.

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100}\times t=2.303lo{g}_{10}\frac{A_0}{A_t}$ ………….(i)

and $\frac{0.693}{50}\times t=2.303lo{g}_{10}\frac{B_0}{B_t}$ ………….(ii)

from equation (i) and (ii)

$0.693t\left[\frac{1}{50}-\frac{1}{100}\right]=\left[log\frac{B_0}{B_t}-log\frac{A_0}{A_t}\right]\times 2.303$

Here; A0 = B0 and $A_t=4\times B_t$

Therefore $0.693t\left[\frac{1}{100}\right]=2.303\left[log\left(\frac{B_0}{B_t}\times \frac{A_t}{A_0}\right)\right]$

$\therefore t=\frac{2.303\times 0.3010\times 2\times 100}{693}=200s$

Here, total meq of acetic acid = 50 × 0.1 = 5

And total meq of NaOH = 25 × 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$pH=pKa+lo{g}_{10}\frac{\left[S\right]}{\left[A\right]}$

$pH=4.76+lo{g}_{10}\frac{2.5}{2.5}=4.76=476\times 10^{-2}$

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