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3 years ago

How many atoms are in 0.00125 moles of zinc

Chemistry

2 answers:

notka56 [123]3 years ago

8 0

Each mole has $6.022 *10^{23}$ atoms.
$\frac{1}{8}* 10^{-2}*6.022* 10^{23} = 7.5* 10^{20}$ atoms

nordsb [41]3 years ago

4 0

One mole of anything is 6.022*10^23 of that thing. so just multiply 6.022*10^23 by 0.00125 to find your answer.

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Which statement is true regarding a triation

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Answer:

It determines the concentration of an unknown substance in neutralization reactions.

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3 years ago

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3 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 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:

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

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 weighted 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

This star of matter has a definite shape and a definite volume

Dominik [7]

Answer:

Solid

Explanation:

7 0

3 years ago

How did Niels Bohr describe electrons in his atomic model? Their energies can have any values. Their exact positions cannot be k

Fed [463]

Answer is: They orbit the central nucleus in discrete paths.

According to the Bohr model of the atom:

1. Electrons orbit the nucleus in orbits that have a set size and energy.

2. Energy levels of electrons are discrete (certain discrete values of energy).

3. Electrons can jump from one energy level to another, absorbing or emitting electromagnetic radiation with a frequency ν (energy difference of the levels).

5 0

3 years ago

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