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2 years ago

What are some non examples of biodiversity

2 answers:

8 0

A monocrop is a non example of biodiversity because it contains only one species, such as all corn, therefore there is very little biodiversity.

777dan777 [17]2 years ago

7 0

Here is the answer
There are multiple ways to look at the concept of biodiversity.
There is diversity on the genetic level within populations, diversity within ecosystems among the species that inhabit them, and even diversity among ecosystems.
<span>You can also look at different scales of biodiversity. For example you can look at a square meter, and see 20 species in it and say that's a pretty diverse square meter. But then, if you look at the neighboring square meter and see the same 20 species, you might begin to question how diverse this ecosystem really is. In contrast, you can look at a square meter of another ecosystem and see 15 species, while its neighboring square meter has a completely different set of 15 species. Which is the ecosystem with greater biodiversity? It all depends on the scale you are interested in.</span>

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which intensive physical property is observed when droppings of a person seated inside a closed room has able to reach a person

zavuch27 [327]

Answer:

The right response will be "Speed".

Explanation:

Intensive or Individualized physical property becomes found when "speed" seems to be the excrement of either an individual who sits within a closed or contained environment that can hit a person one meter away after sneezing or coughing.
This measure seems to be "meter per second", its a intense property.

5 0

2 years ago

A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?

alina1380 [7]

The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!

8 0

2 years ago

Read 2 more answers

A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq)

AnnyKZ [126]

Answer:

$\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}$

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

0.0150 x

$K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}$

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

x 2.293 × 10⁻⁵

$K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}$

4 0

2 years ago

Which of the following statements is true concerning the reaction given below?2H2S(g) + O2(g) → 2S(s) + 2H2O(g)a. The reaction i

pantera1 [17]

Answer:

The rate law may be determined only by experiment.

Explanation:

For a reaction, A + B ---> C, the rate law can only be determined from experimental data. Chemists determine the rate of reaction by carefully observing the changes in the concentration of species as the reaction progresses.

Hence, the rate law is not determined by inspection of the chemical reaction equation, it must be obtained from the experimental data, hence the answer given.

8 0

2 years ago

Three moles of helium gas (molar mass MM = 4.00 g/molg/mol) are in a rigid container that keeps the volume of the gas constant.

Anastaziya [24]

Answer:

The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Explanation:

Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg

The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)

where u = initial rms speed of the gas = 850 m/s

Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J

Then, 3600 J is added to the gas,

New kinetic energy of the gas = 4335 + 3600 = 7935 J

New kinetic energy of Helium atoms = (1/2)(m)(v²)

where v = final rms speed of the gas = ?

7935 = (1/2)(0.012)(v²)

v² = (7935×2)/0.012

v² = 1,322,500

v = 1150 m/s

Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Hope this Helps!!!

8 0

1 year ago