Na₂[Fe(CN)₆]
This compound contains a cation and a complex anion [Fe(CN)₆]^-4
Let the oxidation number of Iron = x
Each cyanide ion has charge = - 1
The sum of the oxidation number of the metal ion and the charges carried'by negative ions should be equal to the charge as the ion.
They x - 6 = - 4 or x = + 2.
Hence, Fe is in +2, oxidation state in this compound.

Molar mass of HNO₃ = 1 + 14 + 3 x 16 = 63
Basicity of HNO₃ = 1
Equivalent mass of HNO₃ = \(\frac { 63 }{ 1 } =63g\) eq^-1

(i) Matter can be classified as solids, liquids and gases based on their physical state.
(ii) Matter can also be classified into mixtures and pure substances based on chemical composition.

One mole is the amount of substance of a system, which contains as many elementary particles as there are atoms in 12 g of carbon -12 isotope.

Molecular mass of silver (Ag) = 107.87 u
Molar mass of Ag = 107.87 g mol^-1
\(\therefore\) Mass of 1 atom of Ag = \(\frac { Molar\ mass }{ Avogadro's\ number } \)
= \(\frac { 107.87g\ { mol }^{ -1 } }{ 6.023\times { 10 }^{ 23 }\ { mol }^{ -1 } } \)
= 17.91 x 10^-23g
Mass of 1 atom of Ag = 17.91 x 10^-23 g.

The solution is a homogeneous mixture of two or more components. Salt in water is homogeneous and therefore it is a solution. Whereas oil in water is heterogeneous or immiscible mixture and so is not a solution

In a redox reaction, the total number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidising agent.

Empirical formula = C₂H₅NO
Molecular mass = 2 x vapour density
= 2 x 29.5 = 59.0
Empirical formula mass of C₂H₅NO = 24 + 5 + 14 +16
= 59
n = \({Molecular \ mass \over Empirical \ formula \ mass }={59\over 59}=1\)
\(\therefore\) Molecular formula = (Empirical formula)_n
= (C₂H₅NO)₁
Molecular formula = C₂H₅NO

Step-1: To find atoms undergoing change in O.N
\(\overset { 0 }{ P } +\overset { +1\quad +5 }{ HNO_{ 3 } } \rightarrow \overset { +1+5-2 }{ HPO_{ 3 } } +\overset { +1-2 }{ NO } +\overset { +1-2 }{ { H }_{ 2 }O } \)
Step-2: To find total decrease and increase in O.N.
P ⟶ HPO₃ (increase in O.N. of 5 units per atom)
HNO₃ ⟶ NO (decrease in O.N. of3 units per atom)
Total decrease 5 x 3 = 15
Total increase 3 x 5 = 15
Step-3: To balance the total increase and decrease in the equation, by multiplying P by 3 and HNO₃ by 5.
3P + 5HNO3  ⟶ HPO₃ + NO + H₂O
Step-4: To balance all atoms other than 'O' and 'H'
3P + 5HNO₃ ⟶ 3HPO₃ + 5NO + H₂O
Step-5: To balance by oxygen atoms
Oxygen and hydrogen atoms balance by themselves.
Hence the balanced equation is 3P + 5HNO3 ⟶ 3HPO₃ + 5 NO + H₂O

Decomposition reaction: Redox reactions in which a compound breaks down into two or more components are called decomposition reactions. These reactions are opposite to combination reactions. In these reactions, the oxidation number of the different elements in the same substance is changed.

Empirical formula = Na₂ SH₂₀ O₁₄ 
\(\left[ \begin{matrix} { Na }_{ 2 }{ SH }_{ 20 }{ O }_{ 14 } \\ =(2\times 23)+(1\times 32)+(20\times 1)+14(16) \\ =46+32+20+234 \\ =322 \end{matrix} \right] \)
n = \(\frac { molar\quad mass }{ caluclated\quad empirical\quad formula\quad mass } =\frac { 322 }{ 322 } =1\)
Molecular formula = Na₂ SH₂₀O₁₄
Since all the hydrogen in the compound are present as water
\(\therefore \) The molecular formula is Na₂ SO₄ 10H₂O.

Empirical Formula = C₆ H₆O
n = \(\frac { molar\ mass }{ calculated\ eprirical\ formula\ mass } \)
= \(\frac { 2\times \ vapour\ density }{ 94 } \frac { 2\times 47 }{ 94 } =1,\) 
Molecular formula (C₆H₆O)  x 1 = C₆H₆O.

mass of one atom  = 6.645  x 10^-23 g
\(\therefore\) mass of 1 mole of an atom = 6.645  x 10^-23 g x 6.022 x 10²³ = 40 g
\(\therefore\) number of moles of element in 0.320 kg = \(\frac { 1\quad mole }{ 40g } \times 0.320kg\)
= \(\frac { 1mol\times 320g }{ 40g } \)
= 8 mol