In this element - Oxygen - the numbers down the right side indicate the ...?

This one's just a simple math problem. Do 88 - 32, which is 56. You can also find the same answer by looking for the atomic mass of iron (Fe), which rounds up to 56.

<h2>Question 2:</h2><h3>This equation is <u><em>balanced</em></u>.</h3>

On the left side, there is 2 Nitrogen (because of the subscript), and 6 Hydrogen (3 x 2 = 6). On the right side, there is 2 Nitrogen (because of the coefficient), and 6 Hydrogen. (2 x 3 again) There are 2 Nitrogen and 6 Hydrogen on each side, therefore the equation is balanced.

<h2>Question 3:</h2><h3>This equation is <em><u>unbalanced.</u></em></h3>

On the left side, there are 2 Nitrogen (N) and 2 Hydrogen (H). On the right side, there is 1 Nitrogen and 6 Hydrogen. There are 2 Nitrogen and 2 Hydrogen on one side, while the other only has 1 Nitrogen and 6 Hydrogen. Therefore the equation is unbalanced.

<h2>Question 4:</h2><h3>This equation is <em><u>balanced.</u></em></h3>

On the left side, there are 2 Chromium (Cr) and 6 Chlorine (Cl). On the right side, there are 2 Chromium and 6 Chlorine. There are 2 Chromium and 6 Chlorine on each side, therefore the equation in balanced.

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If something is confusing or doesn't make any sense, feel free to let me know! ^^

6 0

4 years ago

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

kompoz [17]

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!

4 0

3 years ago

WHICH VIRAL REPRODUCTION CYCLE HAD THE VIRUS PUTTING ITS GENETIC MATERIAL INTO THE CELL

marissa [1.9K]

B lysosgenic hhhhhjjj had to make answer longer

7 0

3 years ago