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Helga [31]
2 years ago
11

Explain the two main physical/ chemical processes by which carbon dioxide molecules in the air move to the cells of phytoplankto

n in the ocean
Chemistry
1 answer:
VARVARA [1.3K]2 years ago
3 0

Answer:

The two physical/ chemical processes by which carbon dioxide molecules in the air move to the cells of phytoplankton in the ocean are the photosynthesis and the biological carbon pump.

Explanation:

The biological carbon pump is the action of organisms to move carbon during chemical and biological interactions from the surface into the deeper ocean and then to rocks.  

The biological carbon pump its composed of three processes, which are the photosynthesis, the gravity and the food web interactions. They are all part of the carbon cycle.

During the photosynthesis, the phytoplankton take up carbon dioxide from the atmosphere that is dissolved in the surface water, and receives the energy from the sun to turn it into glucose and oxygen.  

In the cells of the phytoplankton, glucose is transformed into other organic compounds. This material has organic carbon that can end in two ways: it is incorporated to marine organisms during the food web interactions or it can be remineralised forming calcium carbonate in the ocean surface.  

The remineralization can be done by many organisms to build its shells or skeletons, or by chemical processes that happen in the ocean. This process allows more carbon dioxide to enter the water and to continue the cycle.

So when marine organisms die, all its organic components sink into the bottom of the ocean and carbon-rich sediments are form. And after millions of years, these sediments turn into rocks after going through chemical and physical phenomenon.

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The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell
marysya [2.9K]

Answer:

\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) +   e⁻ ⇌ Ag(s)     E° =  0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)     E° = -0.447    V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

4 0
2 years ago
During a laboratory experiment, 36.12 grams of Al2O3 was formed when O2 reacted with aluminum metal at 280.0 K and 1.4 atm. What
photoshop1234 [79]

Answer:

8.70 liters

Explanation:

4 0
2 years ago
2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta
Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

\eta = \frac{m}{M}

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles  

\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles  

\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles  

\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles  

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles

We have that between MnO₂ and KOH, the limiting reactant is KOH.

\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles

Therefore, the limiting reactant is KOH.

I hope it helps you!

6 0
2 years ago
Read 2 more answers
What is the molarity of a HNO3 solution prepared by adding 290.7 mL of water to 350.0 mL of 12.3 M HNO3?
Lisa [10]

Answer:

6.72M of HNO3

Explanation:

In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:

290.7mL + 350.0mL = 640.7mL

And you are diluting the solution:

640.7mL / 350.0mL = 1.8306 times

As the original concentration was 12.3M, the final concentration will be:

12.3M / 1.8306 =

<h3>6.72M of HNO3</h3>
5 0
2 years ago
Using the data, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 ∘C? (HINT: the act
svp [43]

Explanation:

Below is an attachment containing the solution.

3 0
2 years ago
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