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2 years ago

How does reproduction leads to natural selection in a population

1 answer:

6 0

Natural selection requires variation between individuals. Mutations and reproduction increase genetic variation in a population. Natural selection occurs when environmental pressures favor certain traits that are passed on to offspring.

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PI3Cl2 is a nonpolar molecule. Based on this information, determine the I−P−I bond angle, the Cl−P−Cl bond angle, and the I−P−Cl

alexdok [17]

Answer: Yes I-P-Cl = 90

Explanation:

This is because the angle formed between I-P-Cl is perpendicular hence the angle is 90°

3 0

2 years ago

Which is not a name to describe global winds A) Polar Easterlies B) Prevailing Westerlies C) Trade Winds D) Tropical Northerlies

SVEN [57.7K]

D Tropical northerlies

8 0

3 years ago

Read 2 more answers

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

Mg(OH)2 + HCl --> MgCl2 + H2O

Rashid [163]

I am guessing you want us to balance this equation so.
To balance, we add another molecule of HCl to the left side of the equation and another molecule of water (H20) to the right side of the equation to give:

<span>Mg(OH)2 + 2HCl = MgCl2 + 2H20 </span>

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3 years ago

Of the different states of matter listed below, which is typically the most dense?

stiv31 [10]

The correct answer is option B. The most dense phase of matter is the solid phase and the least dense are gases. However, there is an exception. Water is the exception. Solid water or ice is less dense than the liquid phase therefore it floats on liquid water.

7 0

3 years ago

Read 2 more answers