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dusya [7]
3 years ago
9

In one species of bird, there are three varieties of feather color. What is this an example of?

Chemistry
2 answers:
pishuonlain [190]3 years ago
6 0
Species diversity is the answer
iogann1982 [59]3 years ago
4 0

Answer:

genetic diversity

Explanation:

because its the same species, and color diversity isn't a thing.

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CO2 <br><br> element <br><br><br> number x Atomic mass =total mass
MaRussiya [10]

Answer:

Number= 6

Atomic mass= 12,01115

Explanation:

4 0
3 years ago
What units would be used to measure the diameter of Earth? (site 2)
Masja [62]
We would use miles to measure the diameter of the earth because that is the greatest measure of length that is possible to measure something large.
6 0
3 years ago
Help pls before I end it all
qwelly [4]

Answer:

We might just have to end it together

Explanation:

I tried to answer it now I'm stuck in the same hole -_-

7 0
3 years ago
Read 2 more answers
How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
3 years ago
Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha
nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L

The moles of air are:

3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}

3 0
3 years ago
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