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Ahat [919]
1 year ago
15

N a jungle habitat, there lives a species of snake, called Species A. The carrying capacity of the jungle allows 5000 of Species

A. After their habitat was destroyed, a second species of snakes, Species B, arrives in the jungle. After a few years, the population of Species A has dropped to 0, and the population of Species B has become 5000. What can you determine from this data?
A. They occupy separate niches, but are still competing for some resources.
B. They occupy completely different niches and are not competing.
C. You cannot determine anything from this data.
D. They occupy the same niche and are competing fully for all resources.
Reset Selection
Biology
1 answer:
OLEGan [10]1 year ago
7 0

Based on the carrying capacity of the habitat and the changes that occurred in number of the two snake species, correct option is They occupy the same niche and are competing fully for all resources; option D

<h3>What is carrying capacity of a habitat?</h3>

The carrying capacity of a habitat is the maximum number of species the available resources in that habitat can sustain.

The data given of the species of snakes, Species A and Species B, indicate that they are both present in the habitat after a change occurs in the habitat.

Since the population of Species A became zero after Species B was introduced in the habitat, they occupy the same niche and are competing fully for all resources.

In conclusion, the competition between the species of snakes is responsible for the decline in Species A.

Learn more about carrying capacity at: brainly.com/question/23656166

#SPJ1

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How many grams of urea (non volatile, non electrolyte) must be added to 1250. g water to give a solution with a vapor pressure l
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\large\boxed{\large\boxed{107g}}

Explanation:

The <em>vapor pressure lowering</em> is a colligative property and it follows  Raoult's law.

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               \Delta P=X_{solute}\times P^0

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<u>1. Mole fraction of solute</u>

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           1.39mmHg=X_{solute}\times 55.32mmHg\\\\X_{solute}=0.02513

<u>2. Moles of solute</u>

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  • Xsolvent = moles of solvent / moles of solution

  • 0.97487 = 69.3866mol / moles of solution

  • moles of solution = 71.1752mol

  • moles of solute = moles of solution - moles of solvent = 71.1752mol - 69.3866mol = 1.7886mol

<u>3. Mass of urea</u>

Formula:

  • Mass = number of moles × molar mass
  • Mass = 1.7886 mol × 60.06 g/mol = 107.42 g

You must round to 3 significant figures: 107 g

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