Due to hyperconjugation, alkyl groups act as electron donors when attached to a \(\pi\)-system as shown below:

pH = -log[H⁺] = -log [3.8\(\times \)10^-3]
= - [log[3.8] + log[10^-3]
= - [(0.58) + (-3.0)] = - [-2.42] = 2.42
Therefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic.

The given diagram represents that the represents is carried out in infinite steps, hence, it is isothermal reversible expansion of the ideal gas from pressure 2.0 atm to 1.0 atm at 298 K.
\(W=-2.303 \ nRT \ log\frac { { p }_{ 1 } }{ { p }_{ 2 } }\)
\( W=-2.303\times 1 \ mol\times 8.314 \ { JK }^{ -1 }{ mol }^{ -1 }\times 298 \ K \ log \ 2 \ \ \left( \because \frac { { p }_{ 1 } }{ { p }_{ 2 } } =\frac { 2 }{ 1 } \right)\)
\(w=-1717.63 \ J\)

Moles of O₂, ⁿO₂ = \(\cfrac{mass}{mol.wt.}=\cfrac{8}{32}=0.25mol\)
[mol.wt. of O₂ = 16\(\times\)2 = 32]
Moles of H₂, ⁿH₂ = \(\cfrac{4}{2}=2.0mol\)
[mol.wt. of H₂ = 1 \(\times\)2 = 2 ]
Total number of moles = 0.25 + 2.0 = 2.25mol
Pressure, p = \(\cfrac{nRT}{V}\)
\(\cfrac { 2.25mol\times 0.083bar \ dm^{ 3 }K^{ -1 }mol^{ -1 }\times 300K) }{ (1 \ dm^{ 3 }) } \)
p = 56.025 bar

Nitrogen dioxide, NO₂

As we move from left to right in period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this, effective nuclear charge ivcreases.
More is the effective nuclear charge, more is the attraction between nuclei and electron.
Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character.
Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

a) (n + 1) values are
1s = 1 + 0 = 1, 2s = 2+ 0 = 2, 3s = 3 + 0 = 3, 2p = 2 + 1 = 3
Hence, increasing order of their energy is 1s < 2s < 2p < 3s.
b) 4s = 4 + 0 =4, 3s = 3 + 0 = 3, 3p = 3 + 1 = 4, 4d = 4 + 2 = 6.
Hence 3s < 3p < 4s < 4d.
c) 5p = 5 + 1 = 6, 4d = 4 + 2 = 6, 5d = 5 + 2 = 7, 4f = 4 + 3 = 7, 6s = 6 + 0 = 6.
Hence, 4d < 5p < 6s < 4f < 5d.
d) 5f = 5 + 3 = 8, 6d = 6 + 2 = 8, 7s = 7 + 0 = 7, 7p = 7 + 1 = 8.
Hence, 7s < 5f < 6d < 7p.

There are three common scales to measure temperature ^oC (degree celsius), ^oF(degree Fahrenheit) and K (kelvin). The K is the SI unit.
The temperature on two scales are related to each other by the following relationship \(^{ 0 }F=\frac { 9 }{ 5 } t^{ 0 }C+32\) 
Putting the values in above equation.
\(200-32=\frac { 9 }{ 5 } t^{ 0 }C\Rightarrow \frac { 9 }{ 5 } t^{ 0 }C=168\)   
\(\Rightarrow \ t^{ 0 }C=\frac { 168\times 5 }{ 9 } =93.3^{ 0 }C\)

At high pressure, the real gas shows large deviations from ideal behaviour as the curves are far apart.