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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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A patient gets 2.1 L of fluid over 19 hours through an IV. The drop factor is 20 gtt/mL.Calculate the drip rate in drops per min
Aleks [24]
<span>The rate of infusion is 2.1L/19h or 2100mL/19h (as 1L = 100 mL).
       
To convert 19 hours to minutes we multiply as follows:
   
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8 0
3 years ago
A solution is prepared at that is initially in chloroacetic acid , a weak acid with , and in potassium chloroacetate . Calculate
ratelena [41]

Answer:

2.94

Explanation:

There is some info missing. I think this is the original question.

<em>A solution is prepared at 25 °C that is initially 0.38 M in chloroacetic acid (HCH₂ClCO₂), a weak acid with Ka= 1.3 x 10⁻³, and 0.44 M in sodium chloroacetate (NaCH₂CICO₂). Calculate the pH of the solution. Round your answer to 2 decimal places.</em>

<em />

We have a buffer system formed by a weak acid (HCH₂ClCO₂) and its conjugate base (CH₂CICO₂⁻ coming from NaCH₂CICO₂). We can calculate the pH using the Henderson-Hasselbalch equation.

pH = pKa + log [CH₂CICO₂⁻]/[HCH₂ClCO₂]

pH = -log 1.3 x 10⁻³ + log (0.44 M/0.38 M)

pH = 2.94

4 0
3 years ago
How does a solution become supersaturated?
puteri [66]

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B.)dissolve more solute than you should be able to.

I hope this helps sorry if it doesn't

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