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4 years ago

What process removes carbon dioxide from the ocean?

Chemistry

1 answer:

Katen [24]4 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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The electron configuration of an element is 1s^ 2 2s^ 2 2p^ 2 How many valence electrons does the clement have? (1 point )

Monica [59]

Answer:

Explanation:

8 0

3 years ago

What is the volume of 70.0 g of ether if density os 0.70 g/mL

sdas [7]

$D = m / V$

$0.70 g/mL = \frac{70.0g}{V}$

$V = \frac{m}{D}$

$V = \frac{70.0g}{0.70g/mL}$

$V = 100 mL$

hope this helps!

3 0

3 years ago

50 points - I need this really soon! plz help

kirill115 [55]

Answer:

Explanation:

Butane (C4H10) has 3 C-C bonds in the carbon chain and 10 C-H bonds

7 0

3 years ago

If a climate scientist says, "The CO2 concentration in 2018 was 407.4 ppm." What do they mean? *

Alinara [238K]

Answer:

Ppm means the concentration of vapors or gases expressed in parts per million of polluted air.

Explanation:

In Chemistry and Physics, part per million or PPM is a measure that expresses the number of units of a certain existing substance for every million units of the mixture. (Concentration)

In climate the equation is used so:

ppm = parts of polluted air / million parts of air .

ppm = Liters of polluted air / 10∧6 Liters of air.

This is used to find the concentration of gases expressed as parts of gas per million parts of the

polluted air.

In this case, it would be 407.4L of CO2 per 10∧6L of air.

6 0

3 years ago

During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

??? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0

3 years ago