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4 years ago

Under which condition is competition among fish most likely

Chemistry

2 answers:

belka [17]4 years ago

8 0

Answer:

Answer is: when amount of available plankton is greatly reduced.

Competition is rivalry between two or more organisms for territory, a niche, for scarce resources (in this example available plankon).

Competition occurs naturally between living organisms which co-exist in the same environment (in this example some water area).

Competition between species (in this example fishes) for resources plays a significant role in natural selection.

Genetic variation is important to the population's ability to survive in different situations that affect natural selection. New habitat and increased food supply can increase genetic variation.

seropon [69]4 years ago

3 0

It is under the condition when amount of available plankton is greatly reduced that competition among fish most likely to occur. Hope this answers the question.

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4NH3+5O2=4NO+6H20 if 32.5 grams of NH3 react with enough oxygen, how many grams of water should form?

Aneli [31]

51.3 gram of water should form if 32.5 g of NH₃ react with enough oxygen.

<h3>How to find the Number of moles ?</h3>

To calculate the number of moles use the formula

Number of moles = $\frac{\text{Given Mass}}{\text{Molar Mass}}$

= $\frac{32.5}{17}$

= 1.9 mol

4NH₃ + 5O₂ → 4NO + 6H₂O

4 mol of NH₃ react with oxygen to given 6 mole of water.

So 1.9 mol of NH₃ produces = $\frac{1.9}{4}\times 6$

= 2.85 mol of water

Mass of water = Molar Mass of water × Number of moles of water

= 18 × 2.85

= 51.3 gram

Thus from the above conclusion we can say that 51.3 gram of water should form if 32.5 g of NH₃ react with enough oxygen.

Learn more about the Moles here: brainly.com/question/15356425

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2 years ago

What happens to chemical energy during a chemical change

vivado [14]

It may move around or did never changes

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3 years ago

What is a saturated solution? A. A solution containing more solute than should be possible. B. A solution that can still dissolv

LUCKY_DIMON [66]

Answer:

C. A solution that cannot dissolve any more solute.

Explanation:

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6 0

4 years ago

A car moved 20km East and 70km West. What is the distance?

Nutka1998 [239]

Answer:

50km

Explanation:

70km - 20km = 50 kilometers

6 0

3 years ago

Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate

creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= $\frac{568 g}{142 g/mol}=4 mol$

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= $\frac{2976 g}{32 g/mol}=93 mol$

$2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O$

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

$=\frac{31}{2}\times 4mol=62\text{ mol of}O_2$

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

Oxygen gas is an excessive reagent.
Decane is a limiting reagent.
Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

$=\frac{22}{2}\times 4mol=44\text{mol of }H_2O$

Mass of 44 moles of water ;

$=44mol\times 18g/mol=792g$

792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

brainly.com/question/7144022?referrer=searchResults

3 0

3 years ago