When standardizing an iodine solution, 30.37 ml of the solution was required to completely oxidize 10.00 ml of a 5 mg/ml solutio
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The answers to the two questions are:
1. The percent abundance in the container which has <u>100 navy</u>, <u>27 pinto</u>, and <u>173 black-eyed peas beans</u> is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed <em>peas </em>beans, respectively.
2. The <em>weighted </em>average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.
1. The percent abundance by type of bean is given by:
(1)
Where:
n: is the number of each type of beans
: is the <em>total number</em> of <em>beans </em>
The <em>total number </em>of <em>beans</em> can be calculated by adding the number of all the types of beans:
(2)
Where:
: is the number of navy beans = 100
: is the number of pinto beans = 27
: is the number of black-eyed <em>peas </em>beans = 173
Hence, the total number of beans is (eq 2):
Now, the <em>percent abundance</em> by type of bean is (eq 1):
- Navy beans
- Pinto beans
- Black-eyed peas beans
Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.
2. The average score (S) can be calculated as follows:
(3)
Where:
e: is the score for exams = 85
l: is the score for lab reports = 75
h: is the score for homework = 96
: is the <em>percent</em> for<em> exams</em> = 70.0%
: is the <em>percent </em>for <em>lab reports</em> = 20.0%
: is the <em>percent </em>for <em>homework </em>= 10.0%
Then, the <u>average score</u> is:
We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.
Therefore, the <em>weighted </em>average score will be 84.1.
Find more about percents here:
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Answer: In a compound with molecular formula ,the valency of Y and X are 2 and 3 respectively.
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
The given compound is where X and Y combine and their oxidation states or valencies are exchanged and written in simplest whole number ratios. Thus the oxidation state or valency of X is 3 and that of Y is 2.