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Employers ought to make a clear connection between an applicant's background and experiences and the objectives and requirements of the programme for which they are applying. If the program's emphasis is on project management, for instance, the employer can draw attention to specific examples of the applicant's effective project management, outlining the results and abilities—such as leadership, organization, and problem-solving. The admissions committee can more easily identify how the applicant's professional expertise directly relates to the curriculum because of this alignment.

Although a positive LOR is essential, noting areas for improvement can also provide more legitimacy to the letter. An employer may bring up a certain ability or subject where you first struggled but eventually made progress via work and education. For instance, if you struggled with public speaking at first but overcame your fears by taking classes and practicing, this demonstrates your dedication to self-improvement and your capacity for overcoming obstacles. This well-rounded method offers a more complete picture of your character and career path.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{35}{300}=\frac{40}{T_2}$

$\therefore T_2=\frac{40\times 300}{35}=342.85K$

$T_2\left(°C\right)=69.707\approx 70°C$

$A_2{B}_3?2A^{+3}+3B^{-2}$

123

$\therefore i=1+4\alpha =1+4\times 0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4\times 0.52\times 1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$

Yes, Hygia Institute of Pharmacy admits students based on Class 12, graduation and post graduation scores as per course requirement. The institute admits students based on merit/entrance-exam. Students can check the table below to know the detailed course eligibility:

Maths in VSAT test will include topics from Geometry, Algebra, Trigonometry, Calculus, Probability, and Coordinate Geometry. The questions will be based on Class 11 and 12 NCERT.

$\Delta U=-742.24kJ/mole\Delta H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $-742.24+\frac{1}{2}\times \frac{8.314}{1000}\times 298$

$-742.24+1.238=-741.002\approx -741kJ/mol$

So, magnitude of $\Delta H$ = 741 kJ/mol.

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2\times 9}{\frac{10}{3}\times 2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24\times 10^{-2}=81\times 10^{-2}$

To secure admission at Hygia Institute of Pharmacy, candidates must meet the eligibility criteria set by the college. Aspirants can check the points mentioned below to know more:

1. Aspirants can visit the official website
2. Apply online for admission
3. Students are selected based on merit and entrance-based.

Gabriel phthalimide synthesis is used for 1° Aliphatic /alicyclic amine

K = 3.3 × 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$

3.3 × 10^-4 = $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$  

  $t=\frac{2.303}{3.3}\times 10^4\times 0.22\Rightarrow t=25.58mins$

so; the nearest integer is 26.

Coordination no. of an atom in BCC is 8.

(1) $C{r}^{+1}=4s^{0}3d^{5}M{n}^{+2}=4s^{0}3d^5$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(2) $N{i}^{+2}=4s^{0}3d^{8}C{u}^{+}=4s^{0}3d^{10}$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(3) $F{e}^{+2}=4s^{0}3d^{6}C{o}^{+}=4s^{1}3d^7$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(4) $V^{+2}=4s^{0}3d^{3}C{r}^{+}=4s^{o}3d^5$

The VSAT syllabus is the set of subjects, topics and sub-topics on which the exam question paper will be based.

  $A^2=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$        

$A^4\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^3\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$          

Similarly we get A¹⁹ =   $=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{19}\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right]\&A^{20}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{20}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

=   $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 1+\alpha +\beta =1gives\alpha +\beta =0........\left(i\right)$        

  $2^{20}+\left(2^{19}-2\right)\alpha =4from\left(i\right)$

$\Rightarrow \alpha =\frac{4-2^{20}}{2^{19}-2}=\frac{4\left(1-2^{18}\right)}{-2\left(1-2^{18}\right)}=-2$          

So, b = 2

Hence b - a = 4

Freundlich adsorption Isotherm

$\left(\frac{x}{m}\right)=k{\left(P\right)}^{\frac{1}{n}}$

At moderate pressure ; $\frac{x}{m}$ $\frac{}{}$ varies non-linearly with P.

So, $x=\frac{1}{n}$

$l^{-}+H_2{O}_2+2H^{+}\to l_2+2H_{2}O$

(-1) oxidation (0)

Here, I is reducing agent

${}_{}$ $\therefore H_2{O}_2$  behaves like oxidizing agent

Let the equation of normal is Y – y = - $\frac{1}{m}\left(X-x\right)$  

where m is slope of tangent to the given curve then

$Y-y=-\frac{dx}{dy}\left(X-x\right)$           

It passes through (a, b) so b – y = $\frac{-dx}{dy}\left(a-x\right)$  

->(a – x) dx = (y – b) dy

On integration     $ax-\frac{x^2}{2}=\frac{y^2}{2}-by+c.........\left(i\right)$  

(ii) passes through (3, -3) &  $\left(4,-2\sqrt[]{2}\right)$  then

3a – 3b – c = 9       .(ii)

& 4a -  $2\sqrt[]{2}b$ - c = 12           .(iii)

also given   $a-2\sqrt[]{2}b=3............(iv)$

Solve (ii), (iii) & (iv) b = 0, a = 3

Hence a² + b² + ab = 9

In Ce & Eu (+3) oxidation state is more stable

$\therefore C{e}^{+4}+e^{-}\to C{e}^{+3}\left(Reduction\right)$ so CeO₂ is oxidising agent

$E{u}^{+2}\to E{u}^{+3}+e-\left(oxidation\right)$ so EuSO₄ is reducing agent

The Alliance Manchester Business School has more than 63,000 alumni who are represent over 177 countries and have been making significant contributions to their respective communities. 
 
Mentioned below are some notable members of the B-school's alumni network: 
 
· Sir Terry Leahy: Retired Chief Executive at Tesco 
 
· Simon Collins: Chairman and Senior Partner at KPMG 
 
· Wendy Mayall: Former Chief Investment Officer at Unilever