I have the same question and cant still answer it so I need the answers
This is a bit long right now.
(n l m s) these are the numbers.
n=2, second orbital level.
l is the type of orbit: l=1 elongated, l=0 spherical
m=0, magnetic number: 0
s=spin, both have spin positive
You need still to round it up. Srry!
Answer:
59.8%
Explanation:
First find the Mr of manganese (III) nitrate.
Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>
Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:
Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>
Now we can find percentage composition / percentage by mass of oxygen.
% composition =
× 100
% composition =
× 100 = <u>59.776%</u>
∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).
The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are <span>C) carbon monoxide and carbon dioxide</span>
Element Atomic Number Valency
Valency of Hydrogen 1 1
Valency of Helium 2 0
Valency of Lithium 3 1
Valency of Beryllium 4 2
Valency of Boron 5 3
Valency of Carbon 6 4
Valency of Nitrogen 7 3
Valency of Oxygen 8 2
Valency of Fluorine 9 1
Valency of Neon 10 0
Valency of Sodium (Na) 11 1
Valency of Magnesium (Mg) 12 2
Valency of Aluminium 13 3
Valency of Silicon 14 4
Valency of Phosphorus 15 3
Valency of Sulphur 16 2
Valency of Chlorine 17 1
Valency of Argon 18 0
Valency of Potassium (K) 19 1
Valency of Calcium 20 2
Valency of Scandium 21 3
Valency of Titanium 22 4
Valency of Vanadium 23 5,4
Valency of Chromium 24 2
Valency of Manganese 25 7, 4, 2
Valency of Iron (Fe) 26 2, 3
Valency of Cobalt 27 3, 2
Valency of Nickel 28 2
Valency of Copper (Cu) 29 2, 1
Valency of Zinc 30 2