Class 12th

Amplitude is proportional to the slit – width, thus,

$\frac{A_1}{A_2}=3$            

$\frac{l_{max}}{l_{min}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}=\frac{{\left(3+1\right)}^2}{{\left(3-1\right)}^2}=4.$              

$\alpha =\frac{\Delta l_C}{\Delta l_E}=\frac{3.5}{4}=\frac{7}{8}$  

$\beta =\frac{\alpha }{1-\alpha }=\frac{7/8}{1-7/8}=7$            

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + $\frac{1}{2}r.$

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$

Moles of SO₂ = $\frac{224*10^{-3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18}=2$

Using ; relative lowering in V.P

$\frac{P^o-P}{P^o}=i{x}_B$

Where; $\Delta P=P^0-P$ is lowering in V.P

$\Delta P=P^{0}i{x}_B$

i for NaHSO₃ = 2

here; $x_B=\frac{n_B}{n_A}$ since solution is dilute

$\Delta P=24*2*\frac{0.01}{2}=0.24$

$\Delta P=24*10^{-2}mmHg$

So; x = 24

PV = nRT

1 * V = $\frac{3.12}{32}*0.0821*300$

V = 2.4 litre

Vol of O₂ adsorbed per gm = 2.4 / 1.2 = 2 litre

$C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$

Dichromate ion converted to chromate ion in basic medium and oxidation number of Cr in $Cr{O}_4^{2-}$ is +6.

X → Y

$\begin{array}{l}{E}_{a_f}-E_{a_b}=-20\\ 30-E_{a_b}=-20\end{array}$

$E_{a_b}=50kJ/mole$

$Mn{O}_4^{-}+8H+5e^{-}\to Mn+4H_{2}O{E}^o=1.51V$

Quantity of electricity required to reduce 1 mole of $Mn{O}_4^{-}$ is 5F

So, for 5 mole $Mn{O}_4^{-}$ 25F electricity is required.

[Mn₂ (CO)₁₀]

Bridging ligand (CO) is (0)