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Lesechka [4]
3 years ago
15

Why do higher trophic levels is most ecosystem contain fewer organisms than lower trophic levels

Chemistry
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

With less energy at higher trophic levels, there are usually fewer organisms as well

Explanation: Organisms tend to be larger in size at higher trophic levels, but their smaller numbers result in less biomass. Biomass is the total mass of organisms at a trophic level.

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The accepted value for the molar volume of a gas is
sveta [45]

Answer:

C)10.7%

Explanation:

24.8-22.4=2.4

22.4→100%

2.4→X%

X=2.4×100/22.4=10.7%

7 0
4 years ago
Is NaF an acid, base or salt?
Ymorist [56]

Answer:

NaF is a salt

4 0
3 years ago
Read 2 more answers
If an ultraviolet photon has a wavelength of 77.8 nm calculate the energy of one mole ultraviolet photon.
DerKrebs [107]

Answer:

Explanation:

E = (hc)/(λ)

E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /

(77.8x10^(-9)m)

E = 2.55 x 10^(-11) J

7 0
3 years ago
What units should be used when describing the density of oil?
Alinara [238K]

Answer:

Generally, density is measured using he SI unit for measurement of density which is kg/m³ or the cgs units used to describe the density of a substance which is g/cm³

However, due to its nature (being composed of varying amount proportions of different compounds mixed together resulting in a variation of the density of naturally produced crude oil) of having different densities whereby the lighter oils are more easily processed than the heavier variety of oils, oil density is related to the quality of the oil and it is usually measures in API gravity as follows;

°API = (141.5/S.G.) - 131.5

Where;

API = American Petroleum Institute

S. G. = Specific gravity

As such the API gravity of fresh water with a S. G. of 1.0 has is 10 degrees

Explanation:

6 0
3 years ago
One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

n(\mathrm{O_2}) = n(\mathrm{H}_2).

Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

m = n \cdot M.

In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
  • Right-hand side: \displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9.

Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0
4 years ago
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