Explanation:
Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.
A. One of the oxides (Oxide 1) contains 63.2% of Mn.
Mass of the oxide = 100g
Mass of Mn = 63.2 g
Mass of O = 100 - 63.2
= 36.8 g
Ratio of Mn to O = 63.2/36.8
= 1.72
Another oxide (Oxide 2) contains 77.5% Mn.
Mass of oxide = 100 g
Mass of Mn = 77.5 g
Mass of O = 100 - 77.5
= 22.5 g
Ratio of Mn to O = 77.5/22.5
= 3.44
Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.
B.
Oxide 1
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
Oxide 2
Mass of Mn per 1 g of O = mass of Mn/mass of O
= 77.5/22.5
= 3.44 g/g of Oxygen.
A substance that has no specific volume changes to a substance that has a specific volume.
Answer:
Increasing the surface area of the reactants
Explanation:
An increase in surface area of the reactant will always cause an increase in the rate of reaction. This is so because, an increase in the surface area of the reactant will cause the reactant particles to collide effectively thereby bringing about an increase in the reaction rate.
Collision theory suggests that for a reaction to occur, the reactant particles must collide with the right orientation. As the surface area of the reactants are increased, the reactants particles collide more with the right orientation bringing about definite increase in the rate of reaction.
Answer:
Final Temperature = 36.54 ⁰C
Explanation:
Lets suppose the gas is acting ideally, then according to Charle's Law, "<em>The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature</em>". Mathematically for initial and final states the relation is as follow,
V₁ / T₁ = V₂ / T₂
Data Given;
V₁ = 32 L
T₁ = 10 °C = 283.15 K ∴ K = °C + 273.15
V₂ = 35 L
T₂ = ??
Solving equation for T₂,
T₂ = V₂ × T₁ / V₁
Putting values,
T₂ = (35 L × 283.15 K) ÷ 32 L
T₂ = 309.69 K ∴ ( 36.54 °C )
Result:
As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.
