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Alenkinab [10]
2 years ago
5

why would air moving over a cold current cause fog (advection fog)? group of answer choices the cold current produces the fog wh

en kelp beds release condensation nuclei. all of the other answers are correct, and thus this is the best answer the cold current produces the fog by mixing with the air. the air is chilled, which decreases the capacity of air to hold water vapor, and eve
Chemistry
1 answer:
frutty [35]2 years ago
7 0

The correct answer is option  C.

The air moving over a cold current cause fog or advection fog when the air is chilled, which decreases the capacity of air to hold water vapor, and eventually, the relative humidity reaches 100% - leading to condensation and fog formation.

What is advection fog?

When warm- moist air or warm air front slides over the cold air front or cold surface, it results in the formation of advection fog.

Resultantly, the air becomes saturated and chilled at high humidity levels due to which water vapors start to condense leading to fog formation.

Moreover, the optimal condition for the formation of advection fog is cloudy windy weather having moderate to powerful winds.

If you want to learn more about advection fog click here:

brainly.com/question/18803983

#SPJ4

The complete question is:

Why would air move over a cold current cause fog (advection fog)?

(a)  the cold current produces the fog when kelp beds release condensation nuclei.

(b)  the cold current produces the fog by mixing with the air.

(c) the air is chilled, which decreases the capacity of air to hold water vapor, and eventually, the relative humidity reaches 100% - leading to condensation and fog formation.

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melisa1 [442]

make or become hot or warm is heating.

cooling is to become colder or cool down from a higher temp.

6 0
4 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
When liquid silver nitrate and liquid sodium chloride are combined it becomes
Maru [420]

Answer:

solid silver chloride forms along with a new liquid, sodium nitrate

Explanation:

6 0
3 years ago
Help please you don’t need to show work
aleksandrvk [35]

Answer:

24.07

Explanation:

6 0
3 years ago
How many sig-figs in the following measurements?
DerKrebs [107]

Answer:

1.605cm = 4 significant figures

16.050cm = 4 significant figures

16.050cm = 4 significant figures

12 + 12.5 + 125 = 149.5 = 4 significant figures

1.62 × 10^3/2.8 × 10^-5 = 1620/0.000028 = 11 significant figures

4 0
3 years ago
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