Answer:
The mass percent of aluminum sulfate in the sample is 16.18%.
Explanation:
Mass of the sample = 1.45 g
Mass of the precipitate = 0.107 g
Moles of aluminum hydroxide = 
According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .
Then 0.001372 moles of aluminum hydroxide will be obtained from:
Mass of 0.000686 moles of aluminum sulfate :
= 0.000686 mol × 342 g/mol = 0.2346 g
The mass percent of aluminum sulfate in the sample:
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.
Answer:
is the value of the equilibrium constant at this temperature.
Explanation:
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as 
Partial pressures at equilibrium:
The equilibrium constant in terms of pressures is given as:
is the value of the equilibrium constant at this temperature.
