Answer:Chlorine
Explanation:The distribution of electrons in an atom is called as Electronic Configuration. Formula 2n2 helps in the determination of the maximum number of electrons present in an orbit, here n= orbit number. The formula helps in determination of arrangement of electrons and is known as “Bohr Bury Schemes.”
1. NaCl
2. CaO
3. KOH
4. MgS
5. CuCO3
6. Al2O3
7. Fe2O3
8. Na2CO3
9. AlHO3
10. (NH4)3N
11. Zn3N2
12. MgCO3
All numbers are subscripts.
Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O
With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O
Answer:
Potassium was the first metal to be isolated by electrolysis. It was discovered by the English chemist Sir Humphry Davy by decomposing molten potassium hydroxide (KOH) with a voltaic battery.
Explanation:
Answer:
Silver Acetate would be the Limiting Reagent.
Explanation:
The balance chemical equation for the given double displacement reaction is as;
HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂
Step 1: <u>Calculate Moles of Starting Materials:</u>
Moles of HCl:
Moles = Mass / M.Mass
Moles = 72.9 g / 36.46
Moles = 1.99 moles
Moles of AgC₂H₃O₂:
Moles = 150 g / 166.91 g/mol
Moles = 0.898 moles
Step 2: <u>Find out Limiting reagent as:</u>
According to balance chemical equation.
1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂
So,
1.99 moles of HCl will react with = X moles of AgC₂H₃O₂
Solving for X,
X = 1.99 mol × 1 mol / 1 mol
X = 1.99 mol of AgC₂H₃O₂
Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.
