To solve for the empirical formula, we write first all the data.
Given:
Compound 1: 76 wt% Ru and 24wt% O
Compound 2: 61.2 wt% Ru and 38.8 wt% O
Required: Empirical Formula of Compound 1
Solution:
Assume total mass of the compound is 100 g
Solving for Compound 1,
76 g Ru x <u>1 mol Ru </u> = 0.75195 mol Ru
101.07 g Ru
24 g O x <u>1 mol O </u> = 1.5 mol O
16 g O
Then, divide each mole with the smallest number of moles calculated
Ru = 0.75195 mol/0.75195 mol = 1
O = 1.5 mol/0.75195 mol = 2
Therefore, the empirical formula for Compound 1 is RuO2.
<em>ANSWER: RuO2</em>
In these atoms, the positive and negative charges cancel each other out, leading to an atom with no net charge.
(Protons, neutrons, and electrons)
The atomic number is the number of Protons on the nucleus of the atom, whereas the mass number of some isotope is the mass of the protons plus the mass of the neutrons. That's the difference between isotopes: number of neutrons.
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Answer:
The most active formation of stars found in our galaxy are in the center of milky way near or close to the dwarf planets
