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Furkat [3]
3 years ago
14

Liver cells require a great deal of energy which cell organelles would you except to appear in large number as a result

Chemistry
1 answer:
Arturiano [62]3 years ago
3 0
Mitochondria are the powerhouses of the cell, so mitochondria is the answer
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Generally, does the size of a radioactive sample affect half-life?
Natasha2012 [34]

Answer:

The quantity of radioactive nuclei at any given time will decrease to half as much in one half-life.

Explanation:

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3 years ago
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An electron jumps from energy level 1 to energy level 2 by absorbing a photon of energy 8 eV. What is the most likely energy of
Mkey [24]

Answer:

C. 8 eV

Explanation:

In atoms, the energy of each level is quantised. That is one of the reasons why an electron can not exist between energy levels. Thus, when the electron absorbs energy of 8 ev, it get excited and jumps to energy level 2 which has the current value of energy of the electron.

Thus for the electron to move back to its initial energy level, it must emit the exact energy absorbed for its excitation. Therefore, the likely energy of the photon that will be emitted is 8 ev. This would make the electron to have energy that would make it to exist on energy level 1.

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3 years ago
What is the most accurate sight for a firearm?
Bess [88]

Answer:

a magnifying scope

Explanation: you get a close up view of the target and because of that there is a lower margin of error

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3 years ago
Standard reduction potentials are based on which element?
Andreas93 [3]

Answer: option B. Hydrogen

Explanation:

6 0
4 years ago
There is a mixture of three gases. We know the total pressure is 2.50 atmosphere and the pressure of the oxygen and the nitrogen
Viktor [21]

Answer:

1.14 atm and 1.139 mol

Explanation:

The <em>total pressure</em> of the container is equal to the <u>sum of the partial pressure of the three gasses</u>:

  • P = Poxygen + Pnitrogen + Pcarbon dioxide
  • 2.50 atm = 0.52 + 0.84 + Pcarbon dioxide

Now we <u>solve for the pressure of carbon dioxide</u>:

  • Pcarbon dioxide = 1.14 atm

To c<u>alculate the number of CO₂ moles </u>we use <em>PV=nRT</em>:

  • P = 1.14 atm
  • V = 25.0 L
  • n = ?
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • T = 32 °C ⇒ 32 + 273.16 = 305.16 K

1.14 atm * 25.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305.16 K

  • n = 1.139 mol
5 0
3 years ago
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