Answer:
In pair NaF and H2O both compounds exibit predominantly ionic bonding.
<span>This change is called BOILING.</span>
Hydrogen (H) 1s1
2 Helium (He) 1s2
3 Lithium (Li) [He] 2s1
4 Beryllium (Be) [He] 2s2
5 Boron (B) [He] 2s2 2p1
6 Carbon (C) [He] 2s2 2p2
7 Nitrogen (N) [He] 2s2 2p3
8 Oxygen (O) [He] 2s2 2p4
9 Fluorine (F) [He] 2s2 2p5
10 Neon (Ne) [He] 2s2 2p6
11 Sodium (Na) [Ne] 3s1
12 Magnesium (Mg) [Ne] 3s2
13 Aluminium (Al) [Ne] 3s2 3p1
14 Silicon (Si) [Ne] 3s2 3p2
15 Phosphorus (P) [Ne] 3s2 3p3
16 Sulphur (S) [Ne] 3s2 3p4
17 Chlorine (Cl) [Ne] 3s2 3p5
18 Argon (Ar) [Ne] 3s2 3p6
19 Potassium (K) [Ar] 4s1
20 calcium (ca) [Ar] 4s2
The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Each mole of propane requires 5 moles of oxygen.
Answer:
The answer to your question is: 43 %
Explanation:
Data
NO = 7 mol
O2 = 5 mol
NO2 = 3 mol
percent yield = ?
Reaction
2NO(g) + O2(g) ⇒ 2NO2(g)
Proportion of reactants
From the reaction 2 moles of NO / 1 moles of O2 = 2
From the experiment 7 moles of NO / 5 moles of O2 = 1.4
Then, the limiting reactant is NO.
Rule of three
2 moles of NO -------------- 2 moles of NO2
7 moles of NO -------------- x
x = (7 x 2) / 2
x = 7 moles of NO2
% yield = experimental/ theoretical x 100
% yield = 3/7 x 100
% yield = 42.9 ≈ 43
