Class 12th

One water molecule is associated with hydrogen bond.

Ans. = 1

$L.R.=\frac{\left|3*2+4x-3-5\right|}{\sqrt[]{3^2+4^2}}=\frac{11}{5}$              

Equation of family of parabolas

${\left(x-h\right)}^2=\frac{11}{5}\left(y-k\right)$              

Differentiate 2 (x – h) = $\frac{11}{5}\frac{dy}{dx}$  

Again differentiate 2 = $\frac{11}{5}\frac{d^{2}y}{d{x}^2}$  

$11\frac{d^{2}y}{d{x}^2}=10$              

Given 2x + y – z = 3         . (i)

x – y – z = α        . (ii)

3x + 3y + βz = 3                . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

Please find the below image 

Cl, Br & I form halic (V) acid i.e. HClO₃, HBrO₃ & HlO₃.

If S4 is ranked 6th and S2 is ranked 5th, we have

O > O > O > O > S2 > S4 > O

In this case, S3 will rank the highest and thus S5 will rank the least. We have, S3 > O > O > O > S2 > S4 > S5

Also, the order S7 > S1 > S2 will be maintained The possible arrangements are

S3 > S6 > S7 > S1 > S2 > S4 > S5

or   S3 > S7 > S6 > S1 > S2 > S4 > S5 or      S3 > S7 > S1 > S6 > S2 > S4 > S5

Thus, option (d), i.e., S6 can be ranked third or fourth is true.

If S4 ranks 2nd. S3 ranks 1st and S5 ranks the lowest. The order S7 > S1 > S2 will be followed out of all the given options, only the arrangement given as per option (d) i.e., S3> S4 > S6 > S7 follow. So, option (d) is correct.

If S3 gets the most, S5 ranks the lowest and occupies seventh place. Since S7 always ranks above S1 and S2, . so at the maximum, S1 and S2 will occupy fifth and sixth places. Hence, S7 will not rank lower than fourth.

$\Delta H={\left(E_{act}\right)}_f-{\left(E_{act}\right)}_b$

= x – (x + y) = -y

              = -45 KJ/mol

              Ans. = 45

  $\overrightarrow{b}*\overrightarrow{c}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill \beta \hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -3\hfill \end{array}\right|=\widehat{i}\left(-9-2\beta \right)-\widehat{j}\left(-3+\beta \right)+\widehat{k}\left(5\right)$             

$\left|\overrightarrow{b}*\overrightarrow{c}\right|=5\sqrt[]{3}gives\sqrt[]{{\left(9-2\beta \right)}^2+{\left(\beta -3\right)}^2+25}=5\sqrt[]{3}$         

$81+4{\beta }^2+36\beta +{\beta }^2+9-6\beta +25=75$

$\beta =-2,-4$    

also $\overrightarrow{a}\perp \overrightarrow{b}so\overrightarrow{a}.\overrightarrow{b}=0i.e.1+15+\alpha \beta =0$

So, ${\left|\overrightarrow{a}\right|}^2=1+25+{\alpha }^2=90$