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vodomira [7]
3 years ago
12

Adaptation: Question 1 Which best explains why a population of a certain species of moose , over five generations , developed fe

et that were two times wider than most moose population populations ?
1. The bogs the moose lived in became so soggy that only moose with the largest could successfully walk on the bog .
2 The bogs the moose lived in became so soggy that only moose with the smallest or the largest feet successfully walk on the bogs .
3.The bogs the moose lived in became so soggy that only moose with medium sized feet could successfully walk on the bogs
4. The bogs the moose lived in became so soggy that only moose with the smallest could successfully on the bogs .
Chemistry
1 answer:
slamgirl [31]3 years ago
4 0

Answer:

im doing this test lol

Explanation:

what is it?

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When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
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Answer:

3.73 mL

Explanation:

To solve this problem we can use the equation C₁V₁=C₂V₂, where:

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We<u> input the data</u>:

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And <u><em>solve for V₁</em></u>:

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So 3.73 mL of a 6.57 M stock solution would be required to prepare 500 mL of a 0.0490 M solution.

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How is bond polarity affected by electronegativity?
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Ima Chemist found the density of Freon-11 (CFCl3) to be 5.58 g/L under her experimental conditions. Her measurements showed that
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If we assume this two gases behave like ideal gas, then we use the ideal gas law PV=nRT, where P is the pressure, V is the volume, n number of moles, R gas constant and T the temperature in kelvins. As well as the density formula D=\frac{m}{V}. First we calculate the molarity for Freon

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The number of moles for freon are

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We know n=\frac{m}{M}, so we insert this concept to our ideal gas equation

PV=n \times RT=\frac{m}{M} \times RT= \frac{mRT}{M}. Since D=\frac{m}{V}, we insert this concept to the formula

PV=\frac{mRT}{M} \\P=\frac{mRT}{VM} = \frac{DRT}{M} \\\\\implies M= \frac{DRT}{P}. The measurements take place in same conditions so,

M=\frac{DRT}{P} = \frac{4.38g}{L}\times \frac{1L}{0.0406mol} = 107.9\ g/mol

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