Question
Why is air denser at lower level than at higher altitudes?

Answer

Heavier air comes down and lighter air goes up. Air at lower level on the surface of earth is denser since it is compressed by mass of air above it, that is why it is denser.

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On the basis of the equation $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$, the pH of $10^{-8} \mathrm{~mol} \mathrm{~dm}^{-3}$ solution of HCl should be 8 . However, it is observed to be less than 7.0. Explain the reason.
(a) Name the halide ion whose oxidation can't be done by chemical method. How it can be oxidised?
(b) Explain the bleaching action of chlorine.
Identify (A), (B), (C) and (D) and give their chemical formulae.
$\text{(A)}+\text{NaOH}\xrightarrow{\ \\ \ \ \\ \ \ \\ \ \ }\text{NaCl}+\text{NH}_3+\text{H}_2\text{O}$
$\text{(B)}+\text{NaCl}\xrightarrow{\ \\ \ \ \\ \ \ \\ \ \ }\text{(C)}+\text{NH}_4\text{Cl}$
$\text{(C)}\xrightarrow{\text{Heat}}\text{Na}_2\text{CO}_3+\text{H}_2\text{O}+\text{(D)}$
How many significant figures should be present in the answer of the following calculations?
$\frac{0.02856×298.15×0.112}{0 5785}$
Following cell is set up between copper and silver electrodes.$\text{Cu}|\text{Cu}^{2+}\text{(aq)}\|\text{Ag}^+\text{(aq)}|\text{Ag}$
If its two half cells work under standard conditions, calculate the emf of the cell.
$\Big[\text{Given E}^\circ_{\frac{\text{Cu}^{2+}}{\text{Cu}}}=+0.34\text{V},\text{E}^\circ_{\frac{\text{Ag}^+}{\text{Ag}}}=+0.80\text{V}\Big]$
Write the oxidation and reduction reactions separately from the following redox reaction.
$2\text{Fe}+2\text{H}_2\text{O}+\text{O}_2\xrightarrow{ \ \ \ \\ \ \ }2\text{Fe(OH)}_2$
Give reason for the following statements.
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  2. Electron gain enthalpy of noble gases is almost zero.
An iron rod is immersed in solution containing 1.0M $NiSO_4$ and 1.0M $ZnSO_4.$ Predict giving reasons which of the following reactions is likely to proceed?
  1. Fe reduces $Zn^{2+}$ ions,
  2. Iron reduces $Ni^{2+}$​​​​​​​ ions.
Given,
$\text{E}^\circ_{\frac{\text{Zn}^{2+}}{\text{Zn}}}=-0.76\text{V, E}^{\circ}_{\frac{\text{Fe}^{2+}}{\text{Fe}}}=-0.44\text{V and }$
$\text{E}^\circ_{\frac{\text{Ni}^{2+}}{\text{Ni}}}=-0.25\text{V}$
Complete the following chemical equations:
$\text{Z}+3\text{LiAlH}_4\rightarrow\text{X}+3\text{LiF}+3\text{AlF}_3$
$\text{X}+6\text{H}_2\text{O}\rightarrow\text{Y}+6\text{H}_2$
$3\text{X}+3\text{O}_2\xrightarrow{\triangle}\text{B}_2\text{O}_3+3\text{H}_2\text{O}$
$1 \mathrm{~m}^3$ of $\mathrm{C}_2 \mathrm{H}_4$ at STP Is burnt in oxygen, according to the thermochemical reaction:
$\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \Delta \mathrm{H}=-1410 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
Assuming 70% efficiency, determine how much of useful heat is evolved in the reaction.