v = 2.85 x 10⁸ ms^-1
m_p = 1.673 x 10^-27Kg
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}kgm^{2}s^{-1}}{1.673\times10^{-27}kg\times\times2.85\times10^{8}ms^{-1}}\)
\(\lambda=1.389\times10^{-15}\Rightarrow\lambda=1.389\times10^{-15}A\) \([\because \mathring{A}={10}^{-10}m]\)

Given \(\triangle\)v = 5.7 x 10⁵ ms^-1. \(\triangle\)x = ?
According to Heisenbergs uncertainty principle \(\Delta x \cdot \Delta p \geq \frac{\mathrm{h}}{4 \pi}\)
\( \frac{\mathrm{h}}{4 \pi}=\frac{6.626 \times 10^{-34}}{4 \times 3.14} \mathrm{kgm}^{2} \mathrm{~s}^{-1}=5.28 \times 10^{-35} \)
\(\Delta x \cdot \Delta \mathrm{p} \geq 5.28 \times 10^{-35} \)
\(\Delta x . \mathrm{m} \Delta \mathrm{v} \geq 5.28 \times 10^{-35} \)
\(\Rightarrow \Delta x \geq \frac{5.28 \times 10^{-35} \mathrm{kgm}^{2} \mathrm{~s}^{-1}}{9.1 \times 10^{-31} \mathrm{~kg} \times 5.7 \times 10^{5} \mathrm{~ms}^{-1}} \Rightarrow \Delta x \geq 1.017 \times 10^{-10} \mathrm{~m}\)

(i) The entire quantity of ammonia is consumed in the reaction. So ammonia is the Iimiting reagent. Some quantity of CO₂ remains unreacted, so CO₂ is the excess reagent.
(ii) Quantity of urea formed = number of moles of urea formed x molar mass of urea
= 19 moles x 60 g mol^-1
= 1140 g = 1.14 kg
Excess reagent leftover at the end of the reaction is carbon dioxide.
Amount of carbon dioxide leftover
= number of moles of CO₂ left over x molar mass of CO₂   
= 7 moles x 44 g mol^-1 
= 308 g

The balanced chemical equation is
\(\mathrm{MgCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_{2}\)
Molar mass of MgCO₃ is 84 g mol^–1.
84 g MgCO₃ contain 24 g of Magnesium.
∴ 100 g of MgCO₃ contain

= 28.57 g Mg.
i.e. percentage of magnesium
= 28.57 %.
84 g MgCO₃ contain 12 g of carbon

∴ 100 g MgCO₃ contain

= 14.29 g of carbon.
∴ Percentage of carbon
= 14.29 %.
84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

= 57.14 g of oxygen.
∴ Percentage of oxygen
= 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure MgCO₃ on heating gives 44 g of CO_2.
∴ 1000 g of 90 % pure MgCO₃ gives
\(\frac{\text { wt. of } \mathrm{MgCO}_{3}}{84 \mathrm{~g}} \frac{\text { % Purities }}{100 \%} \frac{\text { wt.of } \mathrm{CO}_{2}}{44 \mathrm{~g}}\)
\(x=44 \times \frac{100}{84} \times \frac{90}{100}\)
= 471.43 g CO₂
= 0.471 kg CO₂

The skeletal equation is:
\(Mn^{3+}_{aq} \rightarrow Mn_{aq}^{2+}+MnO_{2(s)}+H^+_{(aq)}\)
Oxidation half equation:
\(\overset{+3}Mn^{3+}_{(aq)}\rightarrow \overset{+4}MnO_2{s}\)
Balance O.N. by adding electrons,
\(Mn^{3+}_{(aq)}\rightarrow Mn{O}_{2{(s)}}+e^-\)
Balance charge by adding 4H⁺ ions,                   
\(\overset{3+}Mn_{(aq)}\rightarrow MnO_{2{(s)}}+4H^+_{(aq)}+e^-\)
Balance O atoms by adding 2H₂O
\(\overset{ 3+}Mn_{(aq)}+2H_2O_{(1)}\rightarrow MnO_{2{(s)}}+4H^+_{(aq)}+e^-\)....(1)
Reduction half equation:
\(\overset{3+}Mn^{3+}\rightarrow\overset{2+} Mn^{2+}\)
Balance O.N. by adding electrons:
\(Mn^{3+}_{(aq)}+e^-\rightarrow Mn^{2+}_{(aq)}\).....(2)
Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is
\(2M{n}^{3+}_{(aq)}+2{H}_{2}{O}_{(1)}\rightarrow Mn{O}_{2(s)}+M{n}^{2+}_{(aq)}+{4H}^{+}_{(aq)}\)

\(\therefore\) The empirical formula is C₁₄H₉Cl₅.
Calculation of Molecular formula:
The empirical formula mass(C₁₄H₉Cl₅) = \((14\times 12)+(9\times1)+(5\times 35.5)\)
= 168 + 9 + 177.5 = 354.5
\(n=\frac{Molecular\ mass}{Empirical\ formula\ mass}=\frac{354.5}{354.5}=1\)
Molecular formula = (Empirical formula)_n
= (C₁₄H₉Cl₅)_l
\(\therefore\) Molecular formula = C₁₄H₉Cl₅

\(Cr{(OH)}_4^-+H_2O_2\rightarrow Cr{O}_4^{2-}\)
Oxidation half reaction:
\(\underset{(+3)}{Cr{(OH)}_4^-}\rightarrow\underset{(+6)} {Cr{O}_4^{2-}}+3e^-\).... (1)
Reduction half reaction:
\(H_2O_2+e^-\rightarrow H_2O\)....(2)
(+2)                        (+1)

To balance oxygen and hydrogen atoms, OH^- and H₂O are added.
\(2Cr{(OH)}_4^-+3H_2O_2+2OH^-\rightarrow 2Cr{O}_4^{2-}+8H_2O\)

\(S_2O_3^{2-}+I_2\rightarrow S_2O_4^{2-}+SO_2+I^-\)
Oxidation half reaction:

To balance, SO₂ is added on RHS of the equation.
\(S_2O_3^{2-}+I_2\rightarrow S_2O_4^{2-}+2I^-+SO_2\)
To balance oxygen atom, S₂O₃^2- and SO₂ is multiplied by 2.
\(2S_2O_3^{2-}+I_2\rightarrow S_2O_4^{2-}+2I^-+2SO_2\)

Step 2 .4Zn + HNO₃ \(\rightarrow\) 4Zn(NO₃)₂ + NH₄NO₃ + H₂O
Step 3 . To balance N, HNO₃ is multiplied by 10
4Zn + 10HNO₃ \(\rightarrow\) 4Zn(NO₃)₂ + NH₄NO₃ + H₂O
Step 4. To balance oxygen, H₂O is multiplied by 3
4Zn + 10HNO₃ \(\rightarrow\) 4Zn(NO₃)₂ + NH₄NO₃ + 3 H₂O

The variation of the probability of locating the electron on a sphere with nucleus at its centre depends on the azimuthal quantum number of the orbital in which the electron is present.

For 1s orbital, l = 0, m = 0,\(f(\theta)=\frac{1}{\sqrt 2}\)and \(g(\varphi)=\frac{1}{\sqrt 2\pi}\) Therefore, the angular distribution function is equal to \(\frac{1}{2\sqrt \pi }\) 
i.e. it is independent of the angle \(\theta\) and \(\varphi\). Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

For p orbitals, l = 1 and the corresponding m values are -1, 0 and +1. The angular distribution functions are quite complex and are not discussed here, The shape of the p orbital is shown in Figure (b). The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as P_x, P_y and P_z and the angular distribution for these orbitals shows that the lobes are along the x, y arid z axis respectively. As seen in the Figure the 2p orbitals have one nodal plane

For 'd' orbital l = 2 and the corresponding m values are -2, -1, 0, +1, +2. The shape of the d orbital looks like a 'clover leaf'.
The five m values give rise to five d orbitals namely d_xy, d_yz, d_zx, d_x²-y² and d_z². The 3d orbitals contain two nodal planes.

For 'f' orbital, l = 3 and the m values are -3, -2,-1,0, +1, +2, +3 corresponding to seven f orbitals. f_z³, f_xz²,f_yz²,f_xyz,f_{z(x² - y²)},f_{x(x² - 3y²)},f_{y(3x² - y²)} which are shown in Figure. There are 3 nodal planes in the f-orbitals.