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malfutka [58]

Answer:

dwwq is a lot more than a lot more than a year ago and I am a very young ers and I

33rd century fox was a great place to girl and I was a little bit older than a couple of years ago and I was a little bit older than I was a little bit older than I was a little bit older than I was a little bit older than I was a little bit older than I was a little bit older than I was a little bit older than I was a little bit older than I was a little bit

Explanation:

read the full article below to see how the upper c and other factors of a child are at risk for the child or child care act and a lot 5AM in 555inches of a child is filtration of a single person who is filtration

4 0

3 years ago

Which model correctly describes the locations of protons and electrons in the wave mechanical model of the atom?

masya89 [10]

The model correctly describes the locations of protons and electrons in the wave mechanical model of the atom is the proton in nucleas, electron in regions of most probable location.

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5 0

3 years ago

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Which scientist discovered that other galaxies and stars were moving away from us at speeds proportional to how far they were aw

julsineya [31]

I think it was edwin hubble but I'm not sure

7 0

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

A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 3 times the amount needed to compl

Aneli [31]

Answer:

final mole fraction of O₂ = 58.84% , CO₂ = 17.64% , H₂O = 23.52% .

final partial pressure of O₂ = 2.942 atm , CO₂ = 0.882 atm , H₂O = 1.176 atm .

Explanation:

Assuming that propane is present as a gas , and also that the combustion is complete:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

then taking as a reference propane sample= 1 mol , then

initial moles of O₂ = 3* 5 moles = 15 moles

final moles of O₂= 3* 5 moles - 5 moles = 10 moles

final moles of CO₂ = 3 moles

final moles of H₂O = 4 moles

total number of moles = 10 + 3 + 4 = 17 moles

final mole fraction of O₂ = 10/17 = 0.5884 = 58.84%

final mole fraction of CO₂ = 3/17 = 0.1764 = 17.64%

final mole fraction of H₂O = 4/17 = 0.2352 = 23.52%

From Dalton's law for ideal gases , the partial pressure p=P*x then

final partial pressure of O₂ = 5 atm * 10/17 = 2.942 atm

final partial pressure of CO₂ = 5 atm * 3/17 = 0.882 atm

final partial pressure of H₂O = 5 atm * 1/17 = 1.176 atm

4 0

3 years ago

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