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3 years ago

What process removes carbon dioxide from the ocean?

Chemistry

1 answer:

Katen [24]3 years ago

8 0

Answer:

B. decay of dead marine organisms

Explanation:

When the temperature is low, carbon dioxide is captured by the oceans, and when the temperature is high, it is released by the oceans into the atmosphere. At sea, carbon dioxide feeds phytoplankton.

Most of the carbon dioxide consumed by plant plankton (phytoplankton) returns to the atmosphere when this phytoplankton dies or is consumed, but a portion is deposited in the ocean floor sediments when these small particles sink. This process is called a "biological bomb" because carbon dioxide is transported from the atmosphere to the ocean floor.

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Th e enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3. Th e rate of the reaction is

Elis [28]

Answer:

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}$

$V_{max}=k_{cat}[E_o]$

v = rate of formation of products =

[S] = Concatenation of substrate

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$[E_o]$ = Initial concentration of enzyme

We have :

$K_m=0.045 mol/dm^3$

$v=1.15 mmol/dm^3 s=1.15\times 10^{-3} mol/dm^3 s$

$[S]=0.110 mol/dm^3$

$v=V_{max}\times \frac{[S]}{K_m+[S]}$

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

8 0

2 years ago

Given bh3, o3, co2, and nh3, which of these compound is an exception to the octet rules?.

Ghella [55]

Answer:

a. BH₃

Explanation:

According to the octet rules, atoms reach stability when are surrounded by eight electrons in their valence shell when they combine to form a chemical compound.

From the options, the only compound in which the central atom does not meet the octet rules is BH₃. The central atom is boron (B), which has 3 electrons in its valence shell. When B is combined with hydrogen (H), 3 electrons from the 3 atoms of H are added. The total amount of electrons is 6, fewer than 8 electrons needed to meet the rule.

hope this helps

5 0

2 years ago

How many oxygen atoms are present in 30g of glucose?

Alex17521 [72]

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5 0

2 years ago

Can you please help me?

lorasvet [3.4K]

Answer:

60 moles of NaF

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3

From the balanced equation above,

3 moles of NaF reacted to produce 1 mole of AlF3.

Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e

Xmol of NaF = 3 x 20

Xmol of NaF = 60 moles

Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.

4 0

2 years ago

A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper

Anit [1.1K]

Explanation:

Reaction equation for the given chemical reaction is as follows.

$2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}$

Equation for reaction quotient is as follows.

Q = $\frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}$

= $\frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}$

= 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of $SO_{3}$ as equilibrium is achieved by using Q, is as follows.

This means that there are too much products.

Equilibrium will shift to the left towards reactants.

More $SO_{3}$ is formed.

Partial pressure of $SO_{3}$ increases.

4 0

2 years ago