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Leni [432]
4 years ago
14

I wrote this paragraph, and I need help. I need you to write a small, short conclusion. and then I need you to back up my plan w

ith evidence from research. please help. will give brainlyest.
Dear Park Ranger,
The Burmese python is an invasive species in the Everglades ecosystem. This means its numbers increase dramatically due to lack of natural predators. The python is feeding on small mammals such as rodents that feed of egg of animals such as the turtles. This, therefore, causes a ripple effect on the food web in the ecosystem. (ex: dramatically increasing turtle population.)
Due to the lack of natural predators in the ecosystem, the Burmese python causes an imbalance in the7 Everglade ecosystem. They reduce the biodiversity of the ecosystem hence weakening the quality of natural resources.
The number of Burmese pythons can be regulated by introducing a predator in the Everglade ecosystem. Examples of animals that can prey on the snake are eagles, pumas, lions, crocodiles, cobras, and humans. This will reduce the numbers of the Burmese python and restore the balance in the ecosystem.
Chemistry
1 answer:
cluponka [151]4 years ago
3 0

Some people may say the python is not that big of an issue,but when they see how much of a disaster the release of the python can cause,they would probably stop trying to challenge this idea

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How many grams of AlCl3 can be prepared from 3.5 moles of HCl gas?
poizon [28]

Answer:

156 g of AlCl₃ will be produced from 3.5 moles of HCl.

Explanation:

Given data:

Number of moles of HCl = 3.5 mol

Grams of AlCl₃ produced = ?

Solution:

Balanced Chemical equation:

3HCl + Al(OH)₃   →  AlCl₃ + 3H₂O

Now we will compare the moles of AlCl₃  with HCl from balanced chemical equation.

                   HCl          :           AlCl₃

                     3            :              1

                   3.5           :           1/3×3.5 = 1.17 mol

Mass of  AlCl₃ produced:

Mass = number of moles ×molar mass

Mass = 1.17 mol ×   133.341 g/mol

Mass =  156 g

Thus 156 g of AlCl₃ will be produced from 3.5 moles of HCl.

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Answer:

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