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Marizza181 [45]
3 years ago
12

What is the process by which individuals that are better adapted to their evnviroment are more likely to survive and reproduce?

Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
6 0
It is natural selection.
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What volume does 2.25g of nitrogen gas, N2, occupy at 273 Celsius and 1.02 atm​
kotykmax [81]
<h2><u>Answer:</u></h2>

0.126 Liters

<h2><u>Explanation:</u></h2>

V = mRT / mmP

First, convert the 2.25g of Nitrogen gas into moles. (m in the equation above)

2.25g x 1 mole / 28.0g = 0.08036 moles = m

28.0g = mm

Next, convert the 273 Celsius into Kelvin. (T in the equation above)

273 Celsius + 273.15 = 546.15K = T

R = 0.08206L*atm/mol*K

(Quick Note: The R changes depending on the Pressure Unit so do not use this number every time.)

Now, plug everything into the equation.

V = (0.08036)(0.08206)(546.15)/(28.0)(1.02)

V = 0.126 L

5 0
3 years ago
Is relative dating an effective way to determine the age of rocks?
JulsSmile [24]

Answer: Read Explanation

Explanation: Index fossils are useful because they tell the relative ages of the rock layers in which they occur. Geologists use particular types of organisms, such as trilobites, as index fossils.

7 0
3 years ago
Read 2 more answers
Air containing 20.0 mol% water vapor at an initial pressure of 1 atm absolute is cooled in a 1- liter sealed vessel from 200°C t
chubhunter [2.5K]

Answer:

This solution is quite lengthy

Total system = nRT

n was solved to be 0.02575

nH20 = 0.2x0.02575

= 0.00515

Nair = 0.0206

PH20 = 0.19999

Pair = 1-0.19999

= 0.80001

At 15⁰c

Pair = 0.4786atm

I used antoine's equation to get pressure

The pressure = 0.50

2. Moles of water vapor = 0.0007084

Moles of condensed water = 0.0044416

Grams of condensed water = 0.07994

Please refer to attachment. All solution is in there.

6 0
3 years ago
How many grams are 1.20 x 1025 molecules of calcium iodide?
jeka57 [31]

Answer:

1230

Explanation:

1.20×1025=1230 is your answer

5 0
2 years ago
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

5 0
2 years ago
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