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shusha [124]
3 years ago
15

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

iner by type of bean?
2. If your chemistry grade is broken down so that 70.0% of it is based on exams, 20.0% on lab reports and 10.0% on homework, and your average scores (out of 100 points in each area) are: exams: 85, labs: 75, homework: 96. What would your weighted average score be?
Chemistry
1 answer:
kompoz [17]3 years ago
4 0

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:

\% = \frac{n}{n_{t}} \times 100   (1)

Where:

n: is the number of each type of beans

n_{t}: 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:

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 <em>peas </em>beans = 173  

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

n_{t} = 100 + 27 + 173 = 300  

Now, the <em>percent abundance</em> 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 <em>percent</em> for<em> exams</em> = 70.0%

\%_{l}: is the <em>percent </em>for <em>lab reports</em> = 20.0%

\%_{h}: is the <em>percent </em>for <em>homework </em>= 10.0%

Then, the <u>average score</u> is:

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

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:

  • brainly.com/question/255442?referrer=searchResults
  • brainly.com/question/22444616?referrer=searchResults

I hope it helps you!

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A chemist must prepare 300.0mL of nitric acid solution with a pH of 0.70 at 25°C. He will do this in three steps: Fill a 300.0mL
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<u>Explanation:</u>

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Putting values in above equation, we get:

0.70=-\log[H^+]

[H^+]=10^{-0.70}=0.199M

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Molarity of nitric acid = 0.199 M

To calculate the volume of the concentrated solution, we use the equation:

M_1V_1=M_2V_2

where,

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M_2\text{ and }V_2 are the molarity and volume of diluted nitric acid solution

We are given:

M_1=7.0M\\V_1=?mL\\M_2=0.199M\\V_2=350mL

Putting values in above equation, we get:

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2 years ago
How many sodium ions are in 1.4 kg of sodium chloride, NaCl?
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Answer:

1.44 x 10²⁵ ions of Na⁺

Explanation:

Given parameters:

Mass of NaCl  = 1.4kg  = 1400g

Unknown:

Number of ions of sodium  = ?

Solution:

The compound NaCl in ionic form can be written as;

      NaCl →  Na⁺ + Cl⁻

In 1 mole of NaCl we have 1 mole of sodium ions

 Now, let us find the number of moles in NaCl;

  Number of moles  = \frac{mass}{molar mass}  

    Molar mass of NaCl  = 23 + 35.5 = 58.5g/mol

Number of moles  =  \frac{1400}{58.5}     = 23.93mol

 So;

   Since 1 mole of NaCl gives 1 mole of Na⁺  

    In 23.93 mole of NaCl will give 23.93 mole of Na⁺

1 mole of a substance  = 6.02 x 10²³ ions of a substance

  23.93 mole of a substance  =  6.02 x 10²³ x  23.93

                                                   = 1.44 x 10²⁵ ions of Na⁺

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