Answer
The empirical formula is CrO₂Cl₂
Explanation:
Empirical formula is the simplest whole number ratio of an atom present in a compound.
The compound contain, Chromium=33.6%
Chlorine=45.8%
Oxygen=20.6%
And the molar mass of Chromium(Cr)=51.996 g mol.
Chlorine containing molar mass (Cl)= 35.45 g mol.
Oxygen containing molar mass (O)=15.999 g mol.
Step-1
Then,we will get,
Cr=
mol
Cl= 
mol.
O=
mol.
Step-2
Divide the mole value with the smallest number of mole, we will get,
Cr=
Cl=
O=
Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)
Answer:
B. There is a very large percentage of C-12.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.
In such a way, the answer will be B. There is a very large percentage of C-12.
Regards!
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g
<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
- Volume = 250 mL = 250 / 1000 = 0.25 L
- Molarity = 0.3 M
Mole = Molarity x Volume
Mole of Na₂SO₄ = 0.3 × 0.25
Mole of Na₂SO₄ = 0.075 mole
<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
- Molar mass of Na₂SO₄ = 142.05 g/mol
- Mole of Na₂SO₄ = 0.075 mole
Mass = mole × molar mass
Mass of Na₂SO₄ = 0.075 × 142.05
Mass of Na₂SO₄ = 10.65 g
Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.
Learn more about molarity:
brainly.com/question/15370276
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