Given that, formula of first oxide = M₃O₄

Let mass of the metal = x

% of metal in M₃O₄ = (3x / 3x + 64) * 100

but as given % age = (100 - 27.6) = 72.4%

so, (3x / 3x + 64) * 100 = 72.4

or x = 56.

in 2nd oxide,

oxygen = 30%....so metal = 70%

so, the ratio is

M : O

70 / 56 : 30 / 16

1.25 : 1.875

2 : 3

so, 2nd oxide is M₂O₃

5.6lit = 11g

22.4lit = ? (x) g

x = 22.4 × 11 / 5.6

x = 44g

Thus, Molecular mass of gas is 44g

Molar mass of salt = (2 × 23) + 32 + (16 × 4) + x × 18

= 142 + 18x

On heating it will lose water and become anhydrous. 55.9% mass is the mass of water in Na₂SO₄.xH2O.

Mass by mass% = (mass / molar mass of compound) × 100

mass of water = 18x

55.9 = (18x / 142 + 18x) × 100

55.9(142 + 18x) = 1800x

7937.8 + 1006.2x = 1800x

x = 9.99

x = 10 (approx.)

So, the molecular formula of compound is Na₂SO₄.10H₂O.

 Given,

Mass of metal = 0.5 g

Mass of oxide = 0.79 g

First we have to calculate the Mass of Oxygen in metal oxide.

Mass of Oxygen = Mass of oxide - Mass of Metal = 0.79 - 0.5 = 0.29 g

The equivalent weight of oxygen is 8 g/eq.

0.29 g of oxygen combines with the 0.5 g of metal

8 g of oxygen combines with  of metal.

Therefore, the equivalent weight of metal is 13.793 g/eq.

Equation

2 C₂H₂ + 5O₂ → 4CO₂ + 2 H₂O

In this reaction, we can see that, 2 moles of C₂H₂reacts with 5 moles of O_2​ 

Or, 1 mole of ​C₂H₂ reacts with 5 / 2 moles of O_2​

Under standard conditions, we can say that:

22.44 L of C₂H₂ reacts with (5 / 2) x 22.4 L moles of O₂

So, 100 cc of C₂H₂ will reacts with (5 / 2) x 100 cc = 250 cc of O_2​ .

But, since air contains 20 % of oxygen by volume, the amount of air needed to react with 100 cc of C₂H₂ will be = 250 x (100 / 20) = 1250 cc of air.

The equation of the Mn²⁺ in acidic medium  will be.

MnO₄^- + 8H⁺+ 5 Fe²⁺ → Mn²⁺+ 5Fe³⁺ +4H₂O.

We can balance the reaction by using redox balancing of acid and base.

So, we know from the equation that 5 moles of iron react to form 1 mole of Mn⁺².

Thus, we can say that (10*0.1) moles of iron will react with the (10*0.1/5) moles of Manganese ions, which will be 0.2 thus, 10ml volume of KMnO4 and 0.02M moles of KMnO4.  We will get the same result if we use M₁V₁ = M₂V₂ formulae.

Molecular Weight of CCl₄ = 35.5 + 12 x 4 = 154

1 mole CCl₄ vapour=154 g

According to Avogadro’s hypothesis we know that the molar volume of a gas at STP is 22.4 litres

= 154 / 22.4 = 6.87

Dalton's law of multiple proportions is part of the basis for modern atomic theory, along with Joseph Proust's law of definite composition (which states that compounds are formed by defined mass ratios of reacting elements) and the law of conservation of mass that was proposed by Antoine Lavoisier.

The discovery of protons can be attributed to Rutherford. In 1886 Goldstein discovered existence of positively charged rays in the discharge tube by using perforated cathode. These rays were named as anode rays or cannal rays.

Energy is given by formula

E = hc / λ

1 / λ = RZ² [ 1 / n²₁ - 1 / n₂² ]

E = RZ² hc [ 1 / n²₁ - 1 / n²₂]

This energy will be highest when the transition is from n₃ → n₂.