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Paha777 [63]
3 years ago
12

A scientist is wondering why a certain region in the ocean doesn't have maximum phytoplankton growth, despite having plenty of n

utrients. The scientist samples the water for nutrient levels and finds that there is 2.5 ug/L of Phosphorus and 27ug/L of Nitrogen. What can be said about the water?
A) The water is nitrogen limited
B) The water is phosphorus limited
C) The water has an adequate nitrogen and phosphorus, so it is likely iron limited
D) The water has adequate nutrients, but is likely limited by light availability
Chemistry
1 answer:
Advocard [28]3 years ago
8 0

Answer:

C The water had adequate nitrogen and phosphorus, so it is likely iron limited.

Explanation:

Phytoplankton are single- cell organisms that live in oceans.

They require nitrogen, phosphorus and trace amount of iron to survive.

From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.

Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.

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Answer:

Explanation:

hc-- c--ct

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List the number of each type of atom on the right side of the equation hbr(aq)+2naoh(aq)→2nabr(s)+h2o(l)
ser-zykov [4K]

On the basis of the given unbalanced equation, that is:

HBr (aq) + 2NaOH (aq) → 2NaBr (s) + H2O (l)

On the right side of the equation, there are 2 atoms of sodium (Na), 2 atoms of bromine (Br), 2 atoms of hydrogen (H), and 1 atom of oxygen (O₂).

After balancing the equation correctly we get:

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A sample of O2(g) is placed in an otherwise empty, rigid container at 4224 K at an initial pressure of 4.97 atm, where it decomp
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Answer:

The value of K_p at 4224 K is 314.23.

Explanation:

O_2(g)\rightleftharpoons 2O(g)

Initially

4.97 atm            0

At equilibrium

4.97 - p               2p

At initial stage, the partial pressure of oxygen gas = =4.97 atm

At equilibrium, the partial pressure of oxygen gas = p_{O_2}=0.28 atm

So, 4.97 - p = 0.28 atm

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The expression of K_p is given as :

K_p=\frac{(p_{O})^2}{(p_{O_2})}

K_p=\frac{(9.38 atm)^2}{0.28 atm}=314.23

The value of K_p at 4224 K is 314.23.

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ball B

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