Explanation:
from the graph study about oxygen content of Earth's atmosphere, we can understand that
1)
4 billions year ago = None, 3 billions year ago = Cyanobacteria and Archaea , 2 and 1 billions year ago = Bacteria and Green algae , 500 Ma = invertebrate fossils started to existence. Early land plants came in to existence around 398 MA that is Devonian. Dinosaurs are came in to existence during Triassic and Jurassic that is around 251 Ma. Man and animals are recent organism came under Holocene that is 11000 years ago.
2)
The first cells on the earth are anaerobic microorganisms, as the CO2 level is too high they survive by using CO2.
3)
Starting around 2.7 billion years ago, photosynthesis by Cyanobacteria and later plants , pumped “OXYGEN” in to the atmosphere. This caused the decline of anaerobic bacteria and allows the diversification of animals as seen in “CAMBRIAN” around 500 millions year ago.
Early vascular plants “CAPTURED” CO2 starting before the Carboniferous period that began around 350 millions year.Leading to lower temperatures and allowing and allowing the seed plants to outcompetes seedless plants.
Modern human activities has raised both “CO2 and METHANE” level in the atmosphere to over leading to higher temperature and extinction of other species.
Advantages
less wasted energy
Reduces fossilisation
Disadvantages
- Expensive
- doesn't look nice ( apparently)
Answer:
18.0 Ampere is the size of electric current that must flow.
Explanation:
Moles of electron , n = 550 mmol = 0.550 mol
1 mmol = 0.001 mol
Number of electrons = N
![N=N_A\times n](https://tex.z-dn.net/?f=N%3DN_A%5Ctimes%20n)
Charge on N electrons : Q
![Q = N\times 1.602\times 10^{-19} C](https://tex.z-dn.net/?f=Q%20%3D%20N%5Ctimes%201.602%5Ctimes%2010%5E%7B-19%7D%20C)
Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds
1 min = 60 seconds
Size of current : I
![I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BQ%7D%7BT%7D%3D%5Cfrac%7BN%5Ctimes%201.602%5Ctimes%2010%5E%7B-19%7D%20C%7D%7B49.0%5Ctimes%2060%20seconds%7D)
![=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bn%5Ctimes%20N_A%5Ctimes%201.602%5Ctimes%2010%5E%7B-19%7D%20C%7D%7B49.0%5Ctimes%2060%20seconds%7D)
![I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B0.550%20mol%5Ctimes%206.022%5Ctimes%2010%5E%7B23%7D%20mol%5E%7B-1%7D%5Ctimes%201.602%5Ctimes%2010%5E%7B-19%7D%20C%7D%7B49.0%5Ctimes%2060%20seconds%7D%3D18.047%20A%5Capprox%2018.0%20A)
18.0 Ampere is the size of electric current that must flow.
Answer is: <span>an atomic radius.
</span>The atomic radius<span> of a </span>chemical element<span> is a measure of the size of its atom.
</span>The atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.
The atomic radius decreases across the periods because an increasing number of protons, because <span>greater attraction between the protons and electrons.</span>