Chemistry

the lowest spin only magnetic moment for Ni⁺² because of minimum unpaired electron.

$Pb{\left(N{O}_3\right)}_2\underset{673K}{\overset{\Delta }{\to }}PbO+\underset{\left(A\right)}{N{O}_2}+O_2$

$\underset{\left(A\right)}{N{O}_2}\underset{}{\overset{Dimerise}{\to }}\underset{\left(B\right)}{N_2{O}_4}$

$E_{C{\mathcal{l}}_2/C{\mathcal{l}}^{-}}^0=+1.36V$

$E_{l_2/l^{-}}^0=+0.54V$

$E_{A{g}^{+}/Ag}^0=+0.80V$

$E_{N{a}^{+}/Na}^0=-2.71V$

$E_{L{i}^{+}/Li}^0=-3.05V$

$\therefore E_{C{\mathcal{l}}_2/C{\mathcal{l}}^{-}}^0>E_{A{g}^{+}/Ag}^0>E_{N{a}^{+}/Na}^0>E_{L{i}^{+}/Li}^0$

For AB₃ interhalogen compound, which is T-shaped, only two lone pair electrons are available on central atom.

Except electron gain enthalpy all the physical properties like sublimation enthalpy, ionization enthalpy and hydration energy affected the value of reduction potential.

Ans. = 3

Electronic configuration of

Gd is ; Gd₆₄ = [Xe]4f⁷ 5d¹ 6s²

Hence $G{d}_{64}^{++}=\left[Xe\right]4f^{7}5d^{1}6s^0$

Total number of electrons in 4f sub-shell = 7

Ans. = 7

Molecular structure of tetrapeptide Gly-Glu-Asp-Tyr is;

Total negative charge developed at pH 12.5 = 4

Ans. = 4

KCl solution has molality (m) = 3.3, [Means 3.3 mol of KCl dissolved in 1 kg of solved]

Total mass of solution = mass of solute + mass of solvent

= 3.3 * 74.5 + 1000 gm

= 1245.85 gm

Volume of solution = $\frac{Mass}{density}=\frac{1245.85}{1.2}=1038.20ml$

$M=\frac{moles*1000}{V\left(ml\right)}=\frac{3.3*1000}{1038.2}=3.17M$          

Ans. = 3 (the nearest integer)

0.0504 M NH₄Cl of 5ml => millimole of $N{H}_4^{+}=0.0504*5$

0.0210 M NH₃ of 2ml => millimole of NH₃ = 0.0210 * 2

It is a basic buffer.

Total volume = 7ml

$Here,K_b=\frac{\left[O{H}^{-}\right]*\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$     

$\therefore 1.8*10^{-5}=\frac{\left[O{H}^{-}\right]*0.0504*5}{0.0210*2}$         

$\therefore \left[O{H}^{-}\right]=\frac{1.8*10^{-5}*0.0210*2}{0.0504*5}=0.3*10^{-5}M$  

$\therefore \left[O{H}^{-}\right]=3*10^{-6}M$

$\therefore x=3$   

Ans. = 3

If Rate = $K{\left[NO\right]}^a{\left[H_2\right]}^b$  

Case I => $7*10^{-9}=K{\left(8*10^{-5}\right)}^a{\left(8*10^{-5}\right)}^b$

Case II =>   $2.1*10^{-8}=k{\left(24*10^{-5}\right)}^a{\left(8*10^{-5}\right)}^b$

Therefore, $\frac{7\text{?}*10^{-1}}{2.1}={\left(\frac{1}{3}\right)}^a$

or ${\left(\frac{1}{3}\right)}^1={\left(\frac{1}{3}\right)}^a\therefore a=1$           

Ans. = 1