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Sergeeva-Olga [200]
3 years ago
13

What is the limiting factor in determining the accumulation of siliceous ooze/calcareous ooze, respectively?

Chemistry
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

productivity and water depth

Explanation:

The productivity and the depth of water are both equally important as it directly affects the accumulation of biogenic sediments such as the siliceous ooze and calcareous ooze. In the equator and the coastal upwelling areas, and at the site of divergence of oceans, there occurs a high rate and amount of productivity, and these are considered to be the primary productivity.  

The siliceous oozes are a good indicator of extensively high productivity in comparison to the carbonate oozes. The main reason behind this is that the silica can be easily dissolved in the surface water. On the other hand, the carbonates dissolve at a relatively lower ocean water depth, so there requires a high amount of surface productivity in order to allow these siliceous oozes to reach the ocean bottom.

Thus, the water depth and productivity, both are considered as the limiting factor in determining the accumulation of biogenic oozes.

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Which one of the following is most likely to gain electrons when forming an ion, based on the natural tendency of the element?
miskamm [114]

Answer:

Option B. S

Explanation:

All of the options except sulphur, S is metal.

Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.

Sulphur, S has atomic number of 16 with electronic configuration as:

S (16) => 1s² 2s²2p⁶ 3s²3p⁴

From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.

Therefore, sulphur, S will gain two electrons to form ion.

As stated earlier, the rest option given are all metals which will form ion by losing electron(s).

7 0
4 years ago
A certain substance X has a normal freezing point of −3.1°C and a molal freezing point depression constant =Kf·6.23°C·kgmol−1. C
muminat

Answer:

Freezing T° of solution =  - 7.35 °C

Explanation:

This is about the freezing point depression, a colligative property which depends on solute.

The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i

Freezing T° of pure solvent is -3.1°C

At this case, i = 1. As an organic compound the urea does not ionize.

We determine the molality (mol/kg of solvent)

We convert mass to moles:

12.3 g . 1mol / 60.06 g = 0.205 moles

0.205 mol / 0.3 kg = 0.682 mol/kg

We replace data in the formula:

-3.1°C - Freezing T° of solution  = 0.682 mol/kg . 6.23 kg°C/mol . 1

Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C

Freezing T° of solution =  - 7.35 °C

6 0
4 years ago
An atom of an alkaline earth metal reacts with another element to form a compound. Which element MOSTLY LIKELY combined with the
Jobisdone [24]
O sabe esta pregunta
4 0
3 years ago
Read 2 more answers
How much of a 400g sample remains after 4 years if a radioactive isotope has a half-life of 2 years?
timama [110]

Answer:

100 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 400 g

Time (t) = 4 years

Half-life (t½) = 2 years

Amount remaining (N) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Time (t) = 4 years

Half-life (t½) = 2 years

Number of half-lives (n) =?

n = t / t½

n = 4 / 2

n = 2

Thus, 2 half-lives has elapsed.

Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:

Original amount (N₀) = 400 g

Number of half-lives (n) = 2

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 0.25 × 400

N = 100 g

Thus, the amount of the radioactive isotope remaing is the 100 g.

3 0
3 years ago
Jumping on a cemented floor is more pain full than a sandy flour,why​
Anvisha [2.4K]

Answer:

Is soft

Explanation:

because concrete is hard ash so t hink flour would be safer

5 0
3 years ago
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