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Vesna [10]
3 years ago
10

Explain how CO2 is absorbed into the ocean of H2O

Chemistry
1 answer:
Dmitrij [34]3 years ago
7 0
The ocean absorbs CO2 from the atmosphere wherever air meets water. Wind causes waves, giving more opportunity for the water to absorb the carbon dioxide.
You might be interested in
How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

8 0
3 years ago
Which of the following 0.820 M solutions would have the greatest colligative effect?
eimsori [14]

Answer:

K3PO4

Explanation:

Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;

SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)

C4H11N (not ionic in nature hence it can not dissociate into ions)

K3PO4-------> 3K^+ + PO4^3- (4 particles)

Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)

Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.

8 0
3 years ago
Science.... <br> homework!
Leokris [45]
The answer to this question is 2 and 3
5 0
3 years ago
what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?
LuckyWell [14K]
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.
7 0
3 years ago
A positive test with 2,4 dinitrophenylhydrazine could indicate the presence of which of the following functional groups ?
IrinaK [193]
6 &&;$;$//&283$-$-&//
8 0
3 years ago
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