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VikaD [51]
2 years ago
6

Organisms that eat cows do not obtain a great deal of energy from the cows. In fact, they get less energy than the cows obtain f

rom the plants that they eat. Why is this true? A Cows store all their energy in milk. B Cows use energy for their own metabolism. C Cows pass on most of the energy to their offspring. D Cows release all of their energy as heat.
HELP IM TIMED!!!!
Chemistry
1 answer:
Reika [66]2 years ago
8 0

Answer:

<em>The correct option is D) Cows release all of their energy as heat.</em>

Explanation:

Not all of the energy gets travelled from one trophic level to another. Observations have shown that only 10% of the energy travels from one trophic level to another when an organism of the upper trophic level consumes an organism of the lower trophic level. This is because most of the energy is lost by organisms as heat.

So, let's consider that there is 100% energy in plants that the cow eat. The cows will only receive 10% of the energy from the plants. The organisms that will eat the cows will only receive 1%of the energy.

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Which statement accurately compares the trends in atomic number and atomic mass in the periodic table?
Kruka [31]

Answer : The correct option is, (C) Both the atomic mass and the atomic number increase from left to right.

Explanation :

The general trend of atomic number and atomic mass in the periodic table is,

Both atomic number and atomic mass increase from left to right and decreases from right to left in the periodic table due to the addition of the number of neutrons and the number of protons in the nucleus.

Hence, the correct option is, (C) Both the atomic mass and the atomic number increase from left to right.

7 0
2 years ago
Read 2 more answers
Compound w , c6h13cl, undergoes base-promoted e2 elimination to give a single c6h12 alkene, y. compound x, c6h13br, undergoes a
CaHeK987 [17]
<span>Answer: W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation: X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)</span>
5 0
2 years ago
A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the
Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047  mol/dm³

The concentration of the sulfuric acid is 0.047  mol/dm³

Download docx
7 0
1 year ago
From the Bohr equation in the introduction, the calculated energy of an electron in the sixth Bohr orbit of a hydrogen atom is
Natalka [10]

Answer:

<em><u>= - 0.38 eV</u></em>

Explanation:

Using Bohr's equation for the energy of an electron in the nth orbital,

E = -13.6 \frac{Z^{2} }{n^{2} }

Where E = energy level in electron volt (eV)

Z = atomic number of atom

n = principal state

Given that n = 6

⇒ E = -13.6 × \frac{1^{2} }{6^{2} }

<em><u>= - 0.38 eV</u></em>

<em><u></u></em>

<em>Hope this was helpful.</em>

<em><u></u></em>

4 0
2 years ago
Boron has an average mass of 10.81. One isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent. The
Andrej [43]

The average mass of an atom is calculated with the formula:

average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ...  an so on

For the boron we have two isotopes, so the formula will become:

average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)

We plug in the values:

10.81 = 0.1980 × 10.012938  + 0.8020 × mass of isotope (2)

10.81 = 1.98 + 0.8020 × mass of isotope (2)

10.81 - 1.98 = 0.8020 × mass of isotope (2)

8.83 = 0.8020 × mass of isotope (2)

mass of isotope (2) = 8.83 / 0.8020

mass of isotope (2) = 11.009975

mass of isotope (1) = 10.012938 (given by the question)

5 0
2 years ago
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