The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.
<h3>What is an empirical formula?</h3>
It is the minimum ratio between the elements that form a compound.
- Step 1: Divide each percentage by the molar mass of the element.
C: 18.1/12.01 = 1.51
H: 2.27/1.01 = 2.25
Cl: 79.8/35.45 = 2.25
- Step 2: Divide all the numbers by the smallest one.
C; 1.51/1.51 = 1
H: 2.25/1.51 ≈ 1.5
Cl: 2.25/1.51 ≈ 1.5
- Step 3: Multiply all the numbers by 2 so all of them are whole.
C: 1 × 2 = 2
H: 1.5 × 2 = 3
Cl: 1.5 × 2 = 3
The empirical formula is C₂H₃Cl₃.
The empirical formula of the compound with the percent composition C 18.1%, H 2.27%, Cl 79.8% is C₂H₃Cl₃.
Learn more about empirical formula here: brainly.com/question/1603500
#SPJ1
I looked at the question and there isn't a lot of information. im assuming the answer is carbon dioxide. its the only substance that would make sense I hope it helps.
Explanation:
Moles of phosphorus pentachloride present initially = 2.5 mol
Moles of phosphorus trichloride at equilibrium = 0.338 mol
Initially
2.5 mol 0 0
At equilibrium:
(2.5 - x) mol x x
So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol
Mass of 0.338 moles of phosphorus trichloride at equilibrium:
= 0.338 mol × 137.5 g/mol = 46.475 g
Moles of phosphorus pentachloride present at equilibrium :
= (2.5 - 0.338) mol = 2.162 mol
Mass of 2.162 moles of phosphorus pentachloride at equilibrium:
= 2.162 mol × 208.5 g/mol = 450.777 g
Moles of chloride gas present at equilibrium : 0.338 mol
Mass of 0.338 moles of chloride gas at equilibrium:
= 0.338 mol × 71 g/mol = 23.998 g
Answer: hydrogen cleaves from HCl by donating it's only electron to form a radical and chloride ion. Ammonia share it's lone pair of electron with hydrogen to form ammonium ion
Explanation:
