Answer:
Cr₂S₃
Explanation:
From the question given above, the following data were obtained:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Empirical formula =?
Next, we shall determine the mass of sulphur (S) in the compound. This can be obtained as follow:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Mass of sulphur (S) =?
Mass of S = (Mass of chromium sulfide) – (Mass of Cr)
Mass of S = 1.2888 – 0.67
Mass of S = 0.6188 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Cr = 0.67 g
Mass of S = 0.6188 g
Divide by their molar mass
Cr = 0.67 / 52 = 0.013
S = 0.6188 / 32 = 0.019
Divide by the smallest
Cr = 0.013 / 0.013 = 1
S = 0.019 / 0.013 = 1.46
Multiply by 2 to express in whole number
Cr = 1 × 2 = 2
S = 1.46 × 2 = 3
Therefore, the empirical formula of the compound is Cr₂S₃
Answer: Redox Reaction
Explanation:
Redox reaction is the key chemical events in an oxidation-reduction also called Redox. It is the net movement of electrons from one reactants to another
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B. A mixture of the properties of the elements that compose it.
