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Reil [10]
2 years ago
7

What is the physical process whereby atoms or molecules tend to move from an area of higher concentration or pressure to an area

of lower concentration or pressure?
Chemistry
1 answer:
34kurt2 years ago
4 0

Diffusion is the physical process whereby atoms or molecules tend to move from an area of higher concentration or pressure to an area of lower concentration or pressure.

In the field of science, diffusion can be described as a process in which molecules move along the concentration gradient i,e from a region of higher concentration to a region of lower concentration.

The process of diffusion is extremely useful in many everyday living conditions. For example, it is due to diffusion that carbon dioxide and oxygen are exchanged between the lungs and the blood. Diffusion of water, salts, and water is an important process that occurs in the kidneys.

We can divide diffusion into two main types which are simple diffusion and facilitated diffusion. Simple diffusion occurs without any carrier molecule. On the other hand, facilitated diffusion occurs with aid of a carrier molecule.

To learn more about diffusion, click here:

brainly.com/question/94094

#SPJ4

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TRUE OR FALSE<br> “Acids are electrolytes”
Over [174]

Answer:

True mark brainlest

Explanation:

7 0
3 years ago
Read 2 more answers
In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so
alex41 [277]

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: V_{1} = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

M_{1} = 0.321 M,       V_{2} = 21.8 mL = 0.0218 L,      M_{2} = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}

Substitute the values into above formula as follows.

[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

4 0
3 years ago
1. If 500 ML of a 3 L sample of 0.20 M sodium chloride solution is spilled, what is the concentration of the remaining solution?
photoshop1234 [79]
D. 0.2 M

The concentration of a solution is basically the ratio of the solute present to the solvent in the solution. This is an intrinsic property, independent of the amount of solution that is present. A similar example is that of density. No matter the size of a sample, the density and concentration of that sample remain constant.
3 0
3 years ago
Dry air is 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.039% carbon dioxide so varying amounts of water vapor- depending on hu
AfilCa [17]

Answer:

C. Carbon dioxide

Explanation:

Carbon dioxide is one of the end-product of combustion reactions involving many fuels today.

With the rapid increase in urbanization and technological development, man demand for energy increased tremendously. The discovery of fossil fuels paved the way for the astronomical increase in the concentration of carbon dioxide in the atmosphere. The burning of fossil fuels like coal and oil invovles the process where the carbon atoms present in these fuels combine with oxygen in the air to make CO2. This has resulted in an increase in the concentration of atmospheric carbon dioxide (CO2).

The burning fossil fuels for electricity, industry, heat, and transportation are the major sources of the emossion of carbon dioxide.

Also, the cutting down of trees for paper production, building construction and for the establishment of settlements also increase the concentration of carbon dioxide in the atmosphere. Trees are help remove carbon dioxide from the atmosphere through the process of photosynthesis. However, when these trees are cut down, carbon dioxide accumulates in the atmosphere.

5 0
3 years ago
Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

Best regards.

5 0
3 years ago
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