Answer:
marine biology
coastal ecology
astronomy
meteorology
Explanation:
As a college student, to study oceanography one will have to take classes in the field of marine biology, coastal ecology, astronomy and meteorology.
An oceanographer is someone or a professional that studies the ocean in order have more scientific knowledge about them.
Oceanography is a merger of geology, biology, chemistry, physics as it pertains to the ocean.
- There is no need to study human anatomy to study oceanography.
- Marine biology and coastal ecology deals with study of life forms in their marine environment.
- Astronomy and meteorology helps to gain insight about the formation of the ocean and how weather relates to the ocean.
Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:
1. Weigh all the equipment and materials required in the experiment before the experiment.
2. Avoid spillage and evaporation during the experiment.
3. Weigh all the equipment and materials after the experiment.
If the mass is conserved then weight from step 1 is equal to weight from step 3.
There's 6.022×10^23 particles in 1 mole of anything
like there is 1000 grams in 1 kilogram of anything
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
