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Dumas method,

Moles of N in N, N-dimethylaminopentane (C₇H₁₇N)

$=\frac{57.5}{115}=0.5mole$

$C_7{H}_{17}N+\frac{45}{2}CuO\to 7C{O}_2+\frac{17}{2}{H}_{2}O+\frac{1}{2}{N}_2+\frac{45}{2}Cu$

$\frac{n_{CuO}}{\left(\frac{45}{2}\right)}=\frac{n_{c_7{H}_{17}N}}{1}$

$n_{CuO}=\left(\frac{45}{2}\right)\times 0.5mol$

= 11.25 mol

= 1125 × 10^-2 mol

Ans. = 1125

(a) Shows intra molecular H-bonding.

(b) Shows inter molecular H-bonding.

(c) It does not shows intermolecular H-bonding due to high steric hindrance at o-position of benzene ring.

(1) Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

(2) Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

(3) In case of Csl, lattice energy is less but Cs⁺ having less hydration energy due to which it is less soluble in water.

(4) For alkali metal fluorides, the solubility in water increases from Li to Cs. LiF is least soluble in water.

mvr = $\frac{nh}{2\pi }\left(angularmomentum\right)$  

$KE=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$ (Bohr’s kinetic energy)

$=\frac{4h^2}{8{\pi }^{2}m{\left(4a_0\right)}^2}\left(n=2\right)$

$K.E.=\frac{1}{32{\pi }^2}\frac{h^2}{m{a}_0^2}$

Comparing with $\frac{h^2}{xm{a}_0^2}$

x = 32π² = 315.50

10x = 3155

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In the lyophilic colloids, the colloidal particles are extremely solvated.

Let mass of water initially = x gram

Mass of sucrose = (1000 – x) gram

Mole of sucrose = $\frac{1000-x}{342}mol$

$?$ 0.75 molal Þ 0.75 mole solute in 1 kg of solvent

$0.75=\frac{\left(1000-x\right)/342}{x/1000}\left[molality=\frac{n_{solute}}{M_{solvent}\left(kg\right)}\right]$

$4=\frac{\frac{\left(1000-795.86\right)\times 1.86}{342}}{a}$        

a = 0.2775kg or 277.5 gram

Ice separated = (795.86 – 277.5) gram

= 518.3 gram

Ans. = 518 (the nearest integer)

$P=-B\left(\frac{\Delta v}{v}\right)$

or  $B=\frac{-P}{\left(\frac{\Delta v}{v}\right)}$

$B=\frac{-\rho gh}{\left(\frac{-0.005V_0}{V_0}\right)}\Rightarrow h=\frac{908\times 10^8\times 0.005}{10^3\times 9.8}$

$\Rightarrow h=500m$

The median base salary of the MSU Michigan MBA graduates is USD 117,500 for the Class of 2024 (Around INR 1.04 Cr.). Furthermore, as per U.S. News & World Report, the median salary for the graduates is USD 55,084 (Around INR 48.3 lakh). As per the Destination report for the College of Engineering, the median starting salary is USD 77,000 for 2023-24 (Around INR 68 lakh). As per the 2024 overall report for the Bachelor's students, the top employers at this university include Jackson, General Motors, Amazon, PwC, PepsiCo, Deloitte, Trinity Health System, etc.  

$O_2\left(g\right)\underset{}{\overset{UV}{\to }}O\left(g\right)+O\left(g\right)$  

$O\left(g\right)+O_2\left(g\right)\stackrel{UV}{\to }{O}_3\left(g\right)$      

Ozone in the stratosphere is a product of UV radiations acting on O₂ molecule.

$2K_{2}C{r}_2{O}_7+8H_{2}S{O}_4+3C_2{H}_{6}O\to 2C{r}_2{\left(S{O}_4\right)}_3+3C_2{H}_4{O}_2+2K_{2}S{O}_4+11H_{2}O$

Given that, rate of appearance of Cr₂(SO₄)₃ = $+\frac{d\left[C{r}_2{\left(S{O}_4\right)}_3\right]}{dt}=2.67mole/min$  

LHS                                                                 RHS

$\left(\frac{Rateofdisappearanceof{C}_2{H}_{6}O}{3}\right)=\left(\frac{RateofappearanceofC{r}_2{\left(S{O}_4\right)}_3}{2}\right)$            

Rate of disappearance of C₂H₆O = $\frac{2.67\times 3}{2}mol/min$  

Rate of disappearance of C₂H₆O = 4.005 mol /min

Ans. = 4 (nearest integer)

Yes, there is an Indian Students Association at MSU Michigan. This is a culturally registered student organisation committed to fostering inclusivity, equity, and diversity at the university by representing and celebrating the Indian student community. The community of this student organisation includes Undergraduate and Graduate students. It organizes large-scale cultural events such as:

• Diwali
• Holi
• Variety of other celebrations that showcase the richness of Indian heritage

ISO also plays an active role in supporting international students as they transition to life in the U.S.

Some of the subject rankings of MSU Michigan are given below as per some of the top-ranking publishers:

U.S. News & World University Rankings (Global) 2025-26

Times Higher Education (THE) Rankings

Mass of organic compound = 0.2 gram

$n_{AgBr}=\frac{0.188}{188}mol$

= 0.001 mol

$n_{Br}=n_{AgBr}=0.001mole$

 Mass of Br = 0.001 × 80g

= 0.08 g

Mass & of Br = $\frac{0.08}{0.2}\times 100\mathrm{\%}$

= 40 %

According to the top-ranking publisher, MSU ranks among the Top 200 Universities worldwide. The latest rankings of this university are given below:

$2KMn{O}_4\stackrel{\Delta }{\to }\underset{\left(green\right)}{\overset{+6}{K_{2}Mn{O}_4}}+\underset{\left(black\right)}{Mn{O}_2}+O_2$

Mn⁶⁺ has one unpaired electron so paramagnetic and has green colour.

MCl₃.2L => octahedral complex

$MC{l}_3.2L\stackrel{AgN{O}_3\left(excess\right)}{\to }1moleofAgCl$

It means one Cl^- ion present in ionization sphere.

$\therefore \left[MC{l}_2{L}_2\right]Cl$

For octahedral complex co-ordination numbers is 6

$\therefore L$ is bidentate ligand i.e. its denticity is 2.