According to the Law of Definite Proportions from Dalton's Atomic Theory, each compound is composed of a fixed ratio of each of its individual elements. So, the number of individual elements per 1 particle of that compound is represented by the subscripts. The answers are as follows:
Table sugar: 12 atoms of carbon, 22 atoms of hydrogen; 11 atoms of oxygen; 45 total atoms
Marble: 1 atom of calcium, 1 atom of carbon; 3 atoms of oxygen; 5 total atoms
Natural gas: 1 atom of carbon, 4 atoms of hydrogen; 5 total atoms
Rubbing alcohol: 3 atoms of carbon, 8 atoms of hydrogen; 1 atom of oxygen; 12 total atoms
Table sugar: 1 atom of silicon; 2 atoms of oxygen; 3 total atoms
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.
He used that name because if you lit hydrogen on fire it would catch. therefore, it was dangerous to be around. think of it as a kind of warning. :) hope that helps.
Answer:
Explanation:
you would have to look more around the page, for example look at some ways that you can right down.
The density of He is 1.79 x 10⁻⁴ g/mL
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He.
To fill a volume of 6.3 L the mass of He required
= 1.79 x 10⁻⁴ g/mL * 6300 mL
= 11 277 * 10⁻⁴ g
Therefore mass of He required = 1.1277 g of He
