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3 years ago

What happens when a species is moved from its natural environment?

Chemistry

2 answers:

77julia77 [94]3 years ago

3 0

When a species is moved from its natural habitat it will learn to adapt to the new environment

ziro4ka [17]3 years ago

3 0

Answer:

What may happen is everything where they used to live may change. Like The food cycle or the ecosystem. Here's a example. We took wolves from their natural environment and there was a big impact on what they did. Elks is what they ate all the time and kept their population low enough to where the environment can thrive but when they took the wolves the elks started to eat the berries in which Grizzly Bears need to feast on for their hibernation and they ate shrubs and other plant like stuff to which stuff couldn't grow anymore. The Elk population when from 5000 to 30000 from when they took the wolves that day. Another thing is how beavers build dams to keep water from going to certain places which help humans out too, but all the time the Elks would drink water and leave waste inside the water to where fish and other species inside the water started dying out. So beavers couldn't build. So this is my example of what may happen if a species is moved from their natural ground cause they could also ruin a food chain somewhere else.

Explanation:

Hope this helps

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What volume of oxygen gas is released at STP if 10.0 g of potassium chlorate is decomposed? (The molar mass of KClO3 is 122.55 g

ArbitrLikvidat [17]

KClO3(s)+Δ→KCl(s)+32O2(g)

Approx. 3L of dioxygen gas will be evolved.

Explanation:

We assume that the reaction as written proceeds quantitatively.

Moles of KClO3(s) = 10.0⋅g122.55⋅g⋅mol−1 = 0.0816⋅mol

And thus 32×0.0816⋅mol dioxygen are produced, i.e. 0.122⋅mol.

At STP, an Ideal Gas occupies a volume of 22.4⋅L⋅mol−1.

And thus, volume of gas produced = 22.4⋅L⋅mol−1×0.0816⋅mol≅3L

Note that this reaction would not work well without catalysis, typically MnO2.

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Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a

kiruha [24]

From the calculation, the molar mass of the solution is 141 g/mol.

<h3>What is the molar mass?</h3>

We know that;

ΔT = K m i

K = the freezing constant

m = molality of the solution

i = the Van't Hoft factor

The molality of the solution is obtained from;

m = ΔT/K i

m = 3.89/5.12 * 1

m = 0.76 m

Now;

0.76 = 26.7 /MM/0.250

0.76 = 26.7 /0.250MM

0.76 * 0.250MM = 26.7

MM= 26.7/0.76 * 0.250

MM = 141 g/mol

Learn more about molar mass:brainly.com/question/12127540?

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What were some of the “old” ideas that Isaac Newton learned in school?

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Answer:

In addition to mathematics, physics and astronomy, Newton also had an interest in alchemy, mysticism, and theology.

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3 years ago

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1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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3 years ago