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Law Incorporation [45]
3 years ago
13

When glaciers are growing larger, the concentration of the heavier oxygen isotope in sea water increases?

Chemistry
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0

Answer: TRUE

Explanation: Oxygen has two isotopes namely, ¹⁶O (being the lighter oxygen) and ¹⁸O (being the heavier oxygen). Both this isotopes plays a significant role.

The water-ice that accumulates at the glaciated region are derived from the seas and oceans. The ocean and sea water when heated, gets vaporized and rises up, being rich in ¹⁶O, whereas the ¹⁸O, being the heavier isotope does not get vaporize and so it remains in the oceans.

This lighter oxygen isotopes ¹⁶O are then carried to the atmosphere which subsequently falls back as snow in the glacial covered region and consecutively forms ice undergoing compaction.

This, is the reason why ¹⁶O are rich in the glaciated region, whereas the ocean waters are rich in ¹⁸O.

Thus, the above given statement is true, i.e when glaciers become large in size and quantity, then the concentration of heavier oxygen isotope also increases in the ocean and sea water.

Hatshy [7]3 years ago
6 0
Answer : Yes, The heavier oxygen isotope in sea water is increasing as the result of growing of larger glaciers.

The global warming of the climate which is also called as global warming is helping to increase the isotopic oxygen in the ocean and sea waters. When glaciers are growing they have comparatively less isotopic oxygen but this proportion changes with the temperature.
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How many moles of argon are contained in 58 L of At at STP?
Harman [31]

Answer:

n = 2.58 mol

Explanation:

Given data:

Number of moles of argon = ?

Volume occupy = 58 L

Temperature = 273.15 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 58 L = n × 0.0821 atm.L/ mol.K × 273.15 K

58 atm.L = n × 22.43 atm.L/ mol.

n = 58 atm.L / 22.43 atm.L/ mol

n = 2.58 mol

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