Chemistry

The process of moving sodium and potassium ions across the cell membrane is an active transport process involving the hydrolysis of ATP to provide the necessary energy.

The given value of µ (spin only)
2.84 = √n (n + 2) BM, So, n = 2
Among the given configurations, d? system in strong field ligand will have 2-unpaired e? in t? g set of orbitals as shown below.

$\frac{d\left[C\right]}{dt}=\left(\frac{20-10}{10}\right)=1mmold{m}^{-3}$

$+\frac{d\left[D\right]}{dt}=\left\{-\frac{d\left(B\right)}{dt}\right\}*1.5=\frac{3}{2}\left\{-\frac{d\left[B\right]}{dt}\right\}$

$\left\{-\frac{d\left[B\right]}{dt}\right\}=2*\left\{\frac{-d\left[A\right]}{dt}\right\}$

$\therefore \frac{1}{6}\left\{\frac{-d\left[B\right]}{dt}\right\}=\frac{1}{3}\left\{\frac{-d\left[A\right]}{dt}\right\}=\frac{1}{9}\left\{\frac{+d\left[D\right]}{dt}\right\}=\frac{d\left[C\right]}{dt}$

$\therefore$ rate of reaction = $\frac{+d\left[C\right]}{dt}$  = 1m.m dm^-3 S^-1

4-ethoxycarbonylpent-3-enoic acid

$F{e}^{+3}+l^{-}\to I_2+F{e}^{+2}$

Fe⁺² undergoes reduction & I₂ undergoes oxidation.

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

= (0.77 – 0.54) V = 0.23 V

 = 23 * 10^-2 V

$\therefore$  x = 23

 x = 23

In S?2 reaction, inversion takes place, so configuration get reversed.

$N_2{O}_4?2N{O}_2$

$\underset{=0.5mol}{\left(1-0.5\right)}$ $2*0.5=1mol$

$k_P=\frac{{\left(\frac{1}{1.5}*1\right)}^2}{\left(\frac{0.5}{1.5}*1\right)}$

$=\frac{4}{3}=1.33$

$=-710.15J/mol$

As Zn (OH)? , BeO, Al? O? are amphoteric, so they can react with both HCl and NaOH.

When Ferric chloride reacts with potassium thiocyanate a blood red colour of Ferric thiocyanate is formed
FeCl? + 3KSCN → Fe (SCN)? + 3KCl

1000 ml solution contains 0.02 milli mole (mm)

$\therefore$ 500 ml solution contains 0.02 m.m

$\therefore$ Solution made 1000 ml with H₂O

$\therefore$ m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H₂SO₄

= 0.01 + 0.01

= 0.02 m.m

                             = 0.00002 * 10³ mm