<u>Answer:</u> The empirical and molecular formula of the compound is and respectively
<u>Explanation:</u>
We are given:
Mass of C = 3.758 g
Mass of H = 0.316 g
Mass of O = 1.251 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen =
Moles of Oxygen =
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.
For Carbon =
For Hydrogen =
For Oxygen =
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of C : H : O = 4 : 4 : 1
The empirical formula for the given compound is
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is:
We are given:
Mass of molecular formula = 130 g/mol
Mass of empirical formula = 68 g/mol
Putting values in above equation, we get:
Multiplying this valency by the subscript of every element of empirical formula, we get:
Hence, the empirical and molecular formula of the compound is and respectively
Answer:
Explanation:
Fe⁺²(aq) + ClO₂(aq) → Fe⁺³(aq) + ClO₂⁻(aq)
Here oxidation number of Fe is increased from +2 to +3 , so Fe is oxidised .
The oxidation number of Cl is reduced from + 4 to +3 so Cl is reduced .
So ClO₂(aq) is oxidising agent and Fe⁺²(aq) is reducing agent .
Answer: The concentrations of at equilibrium is 0.023 M
Explanation:
Moles of =
Volume of solution = 1 L
Initial concentration of =
The given balanced equilibrium reaction is,
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of at equilibrium is 0.023 M
<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.
So the answer is:
8,12,48
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