You start by finding the mol of each
59.9g C x (mol C / 12.01 g C) = 4.98 mol C
8.06g H x (mol H / 1.00 g H) = 8.06 mol H
32.0 g O x (mol O / 16.0 g O) = 2 mol O
So when you set it up you have
C4.98 H8 O2
You divide each by the smallest mol. The smallest mol is 2
C2 H4 O2.5
However you can’t have half a mol in the empirical formula. If the value ends in 0.5, you multiply everything by 2
You’re left with
C2H8O5
The EMPIRICAL formula for lucite is C2H8O5
Note empirical is not the same as chemical formula.
Answer:
No of neutrons and protons decides mass of the atom. Electrons having negligible
mass are ignored.
Mass of atom
Mass of atom= sum of no of protons(Z) and neutons(N)
=Z+N
For eg:
In hydrogen atom,
Mass of H= Z+N
=1+0
=1 AMU
Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V
If an excess of phosphate enters the waterway, algae, and aquatic plants will grow wildly, choke up the waterway and use up large amounts of oxygen.
