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2 years ago

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

Chemistry

1 answer:

zhannawk [14.2K]2 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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The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

Elden [556K]

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

6 0

2 years ago

Questioned document examiners prefer to work with copies of a questioned document in order to avoid damage to the original

gladu [14]

I think it is a False

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2 years ago

How many moles of O2 are needed to react completely with 35.0 mol of 10 points

yKpoI14uk [10]

D.10.0 mol is the correct

4 0

2 years ago

Express 6009m in scientific notation

Andre45 [30]

6.009 x 10^3. You have to put 6009 into a decimal that is less than 10. Count backwards from 9 until you get to a single digit. You would move the decimal back three places making it 10^3

4 0

2 years ago

Read 2 more answers

Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) an

mafiozo [28]

Answer:

$\rm K$ : approximately $22.0\%$ .

$\rm Cl$ : approximately $19.9\%$ .

$\rm O$ : approximately $27.0\%$ .

$\rm Al$ : $31.0\%$ .

Explanation:

Consider a $100\; \rm g$ sample. There would be $69.0\% \times 100\; \rm g = 69.0\; \rm g$ of $\rm KClO_{3}$ and $100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g$ of $\rm Al$ .

Out of that $69.0\; \rm g$ of $\rm KClO_{3}$ :

$31.9\%$ (by mass) would be $\rm K$ : $31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g$ .

$28.9\%$ (by mass) would be $\rm Cl$ : $28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g$ .

$39.2\%$ (by mass) would be $\rm O$ : $39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g$ .

Overall, the composition (by mass) of each element in this mixture would be:

$\rm K$ :

$\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}$ .

$\rm Cl$ :

$\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}$ .

$\rm O$ :

$\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}$ .

$\rm Al$ :

$\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}$ .

7 0

2 years ago