Answer:
Keep it simple. If all the oxygen contained in the 200 grams of potassium chlorate is produced in the decomposition, then all we have to do is find out how many grams of oxygen are there in the 200 grams. This we can do by calculating the ratio of oxygen mass to the whole. Using 39.1 for potassium, 35.45 for chlorine and 3 times 16, or 48 for the oxygen, we get a total of 122.55 grams per mole for potassium chlorate, of which 48 grams are oxygen. This ratio is 48/122.55. This ratio times the original 200 grams of the compound, gives us 78.34 grams of oxygen produced.
Explanation:
Explanation:
Some Rules Regarding Oxidation Numbers:
- Hydrogen has oxidation number of + 1 except in hydrides where it is -1
- Oxygen has oxidation number of -2 except in peroxides where it is -1
- Some elements have fixed oxidation numbers. E.g Halogen group elements has oxidation number of -1
- Oxidation number of a compound is the sum total of the individual elements and a neutral compound has oxidation number of 0.
A. HI
Hydrogen has oxidation of + 1
Oxidation number of I:
1 + x = 0
x = -1
B. PBr3
Br has oxidation number of - 1
Oxidation number of Pb:
x + 3 (-1) = 0
x = + 3
C. KH
Hydrogen has oxidation of + 1
Oxidation number of K:
1 + x = 0
x = -1
D. H3PO4
Hydrogen has oxidation number of + 1
Oxygen has oxidation number of -2
Oxidation number of P:
3(1) + x + 4(-2) = 0
3 + x - 8 =0
x = 5
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
