They are always univalent because after losing one electron, they aquire nearest inert gas configuration.Li⁺ forms largest hydrated cations because it has the highest hydration energy.

chlorine is an oxidant and K₂MnO₄ is reductant.

The elements of s and p-block are collectively called representative or main group elements. These include elements of group I (alkali metals), group 2 (alkaline earth metals).

44.8 litres of sulphur dioxide at N. T.P.
22.4 litres of sulphur dioxide at N.T.P. = 1 mol
44.8 litres of sulphur dioxide at N.T.P. =\(\frac { 1 \ mol }{ (22.4 \ L) } \) x (44.8L) = 2.0 mol

(i) \({ CH }_{ 2 }={ CH }_{ 2 },CH_{ 3 }CH={ CH }_{ 2 },CH_{ 3 }{ CH }_{ 2 }CH={ CH }_{ 2 },CH_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }CH={ CH }_{ 2 }\)
(ii) \(HCOOH,{ CH }_{ 3 }COOH,{ CH }_{ 3 }{ CH }_{ 2 }COOH,{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }COOH\)
(iii) \({ CH }_{ 3 }CO{ CH }_{ 3 },{ CH }_{ 3 }{ CH }_{ 2 }CO{ CH }_{ 3 },{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }CO{ CH }_{ 3 },{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }CO{ CH }_{ 3 }\)
(iv) \({ CH }_{ 3 }OH,{ CH }_{ 3 }{ CH }_{ 2 }OH,{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }OH,{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }OH\)
(v) \(\\ HC\equiv CH,{ CH }_{ 3 }C\equiv CH,{ CH }_{ 3 }{ CH }_{ 2 }C\equiv CH,{ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }\equiv CH\)

q = 2.303 nRT log \(\frac { { V }_{ 2 } }{ V_{ 2 } } \) = 2.303\(\times \frac{10}{40}\times log\frac{5}{10}\)
=-103.635 cal
For isothermal expansion, \(\Delta\)U = 0  ...........(i)
W = \(\Delta\)U - q = 0 - (-103.635)
= +103.635 cal 
Also, when temperature is constant,  
p₁V₁ = p₂V₂ or pV = constant
\(\Delta\)H = \(\Delta\)E + \(\Delta\)(pV) = 0 + 0 = 0 ..........(ii)

N₂   \(\Longleftrightarrow \)  2N
Initial moles
\(\cfrac{1.4}{28}=0.05\)
Final moles   
\(0.05-\cfrac{30}{100}\times0.05 \ 2\times0.015\)
Dissociation = 0.05 - 0.015 = 0.035 = 0.030
\(\therefore\) Total number of moles = 0.035 + 0.030 = 0.065
i.e. n = 0.065 mol, V = 5 L, T = 1800 K, p = ?
From pV = nRT, we get
p = \(\cfrac{nRT}{V}\)
= 1.92 atm

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.
\(200-32=\frac { 9 }{ 5 } t^{ 0 }C\Rightarrow \frac { 9 }{ 5 } t^{ 0 }C=168\)   
\(\Rightarrow \ t^{ 0 }C=\frac { 168\times 5 }{ 9 } =93.3^{ 0 }C\)

It is defined as the ratio of product of molar concentration of products to the product of molar concentration of reactants at any stage of reaction.
Q_c = \(\frac { \left[ C \right] ^{ c }\left[ D \right] ^{ d } }{ \left[ A \right] ^{ a }\left[ B \right] ^{ b } } \)
for the reaction 
aA + bB ⇌ cC + dD

It is a U-shaped tube filled with agar-agar containing inert electrolyte like KCl or KN0₃ which does not react with solutions

CH₄ = sp³
CH = sp²
C₂H₂ = sp

Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles.
Therefore, it undergoes electrophilic substitution reactions very easily.

Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benzene undergoes nucleophilic substitutions with difficulty.

Solid carbon dioxide is known as dry ice.It is soft, white snow like substances and looks like ice.However, it does not wet a piece of cloth or paper because it sublimes without melting.Solid carbon dioxide is used as a refrigerant under the commercial name of drikold.

Superoxides react with water to form H₂O₂ and O₂.
2\({ O }_{ 2 }^{ - }\)  + 2 H₂ O  \(\longrightarrow \) 2 OH^- + H₂ O₂ + O₂

\( {[K O H]=\frac{0.561}{56} \times \frac{1000}{200} M=0.050 M} \)
\( \text { As } K O H \rightarrow K^{\oplus}+\stackrel{\ominus}{O} H, \therefore\left[K^{\oplus}\right]=[\stackrel{\ominus}{O H}]=0.05 M \)
\( {\left[H^{\oplus}\right]=K_w /\left[{ }^{\Theta} \mathrm{O} H\right]=10^{-14} / 0.05=10^{-14}} \)
\(\left(5 \times 10^{-2}\right) \)
\(=2.0 \times 10^{-13} M, p H=12.7\)
pH = 12.7

No, the oxidation number of an element in any species is an apparent charge on the atom which it appears to have acquired when all other atoms in the species are removed as ions.

Q = \(\Delta U\) + w
w = 430 on system
Heat lost (Q) = -120
\(\Delta U\) = q - w
= -120 - 430
= +310 J

From the information and data given in the question, we concluded that Hydrogen fluoride has highest dipole moment is due to highest electronegativity of fluorine as well as due to the presence of hydrogen bonding in HF. Therefore, HF has highest boiling point.

Charge on the oil drop = 1.282 ×10^–18C
Charge on one electron = 1.6022 × 10^–19C
∴ Number of electrons present on the oil drop
\(=\frac{−1.282 × 10^{−18}C}{−1.6022×10^{−19}C}\)=0.800×10=8.0 electrons

One mole of any substance contains 6.022\( \times\)10²³ atoms /molecules
Hence, number of millimoles of H₂SO₄
= molarity \( \times\) volume in mL
= 0.02 \( \times\)100
= 2millimoles
=  2\( \times\) 10^-3 mol 
Number of molecules = number of moles \( \times\) N_A  
= 2\( \times\)10^-3\( \times\)6.022\( \times\)10²³
=12.044 x 10²⁰ molecules

Sodium bicarbonate (NaHCO₃)  on hetaing composes to produces CO₂ gas which whwn pases thrugh lime water, turns milky.
2NaHCO₃ \(\xrightarrow[]{\Delta}\)Na₂CO₃ + CO₂ + H₂O
Ca(OH)₂(aq) + CO₂(s)\(\rightarrow \)CaCO₃(s) + H₂O(l)
But Na₂CO₃ (Sodium carbonate is stable to heat does not decompose to heat).

Calculation of molecular formula
Empirical formula mass =
 \(2\times 56+3\times 16=160\)
\(n=\frac { molecular \ formula }{ empirical \ formula \ mass } =\frac { 160 }{ 160 } =1\)
\(\therefore Molecular\ formula={ A }_{ 2 }{ B }_{ 3 }\)