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prohojiy [21]
4 years ago
8

There is a correlation between algae height and the snout-to-vent length of a group of marine iguanas in the Galapagos; as algae

height increases for a given island, so does iguana length. Why would the body size of this population of iguanas correlate with algae height?
Differential reproduction of larger animals is favored when adequate nutrition is supplied by larger algae.
Increased variation in animal size is favored when large plant species are present in a healthy ecosystem.
Overproduction of smaller animals is favored when algae swards are short and vice versa when algae swards are long.
Traits are exchanged between iguana and algae populations so that taller algae result in larger iguana populations
Chemistry
2 answers:
beks73 [17]4 years ago
6 0

Answer;

The correct answer would be Differential reproduction of larger animals is favored when adequate nutrition is supplied by larger algae.

The size of the food influences the size of the organism.

Due to availability of larger food or larger algae, longer iguanas tend to live and survive more.

Thus, the tend to survive and reproduce more as compared to the shorter iguanas.

Similarly, when algae or food is not available, then larger iguanas die due to malnutrition and shorter iguanas tend to survive.

Anna11 [10]4 years ago
3 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

The body size of this population of iguanas correlate with algae height because <span>Traits are exchanged between iguana and algae populations so that taller algae result in larger iguana populations</span>
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nika2105 [10]

Answer: The molecular formula will be H_{16}NOCl

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of H = 5.80 g

Mass of N = 20.16 g

Mass of O = 23.02 g

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Step 1 : convert given masses into moles.

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.80g}{1g/mole}=5.80moles

Moles of N =\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{20.16g}{14g/mole}=1.44moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.02g}{16g/mole}=1.44moles

Moles of Cl =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{51.02g}{35.5g/mole}=1.44moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For H = \frac{5.80}{1.44}=4

For N = \frac{1.44}{1.44}=1

For O = \frac{1.44}{1.44}=1

For Cl = \frac{1.44}{1.44}=1

The ratio of H: N: O: Cl= 4: 1: 1: 1

Hence the empirical formula is H_4NOCl

The empirical weight of H_4NOCl = 4(1)+1(14)+ 1(16) + 1(35.5)= 69.5 g.

The molecular weight = 278 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{278}{69.5}=4

The molecular formula will be=4\times H_4NOCl=H_{16}NOCl

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Answer:

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What is the name of the functional group that is attached to this hydrocarbon? The first and last of a chain of three carbons ar
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Answer:

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Explanation:

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