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4 years ago

A certain blue solid contains 36.84% n and 63.16% o. what is the empirical formula of this compound?

1 answer:

3 0

Divide the percentages by the RAM and then divide by the smallest value, this will give a ratio of the elements in the compound. N =14, O=16
Nitrogen : Oxygen
36.84/14 : 63.16/16
2.5 : 3.9
Divide both values by the lowest, and try to find whole number ratios:
2.5/2.5 : 3.9/2.5
1 : 1.5
Since the ratio of oxygen to nitrogen is 1.5 : 1, we have to double it so that it become 3:2
The empirical formula would be N₂O₃

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The answers to the two questions are:

1. The percent abundance in the container which has 100 navy, 27 pinto, and 173 black-eyed peas beans is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The weighted 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:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the total number of beans

The total number of beans can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed peas beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the percent abundance by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

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:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the percent for exams = 70.0%

$\%_{l}$ : is the percent for lab reports = 20.0%

$\%_{h}$ : is the percent for homework = 10.0%

Then, the average score is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

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