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Yuki888 [10]
3 years ago
7

How does the amount of humidity affect hurricane formation?

Chemistry
1 answer:
svetoff [14.1K]3 years ago
8 0
<span>A Hurricane needs humidity to develop. A hurricane starts off as a series of thunderstorms which intensify as it moves over the (warm) humid sea.</span>
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Formula empirica del glutamato monosódico
Citrus2011 [14]

Answer:

C₅H₈NO₄Na

Explanation:

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Alpha Centauri B is a star that is about the same size as the Sun. Why does it appear much smaller and dimmer than the Sun when
Ugo [173]
 To the eye – taken by itself – Alpha Centauri A appears as the fourth-brightest star seen from Earth, just slightly outshone by Arcturus. However, the combined light of Alpha Centauri A and B is slightly greater than Arcturus, so in that sense it appears as the third-brightest star visible in Earth's sky.
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Read 2 more answers
HELP PLZ
Vaselesa [24]

Answer:

Scientific Explanation of weather change and cause storms is described below in detail.

Explanation:

The weather is just the nature of the environment at any time, including items such as warmth, rainfall, air pressure, and cloud shelter. Daily variations in the weather are due to storms and winds. Seasonal variations are due to the Earth spinning nearby the sun. Warm beginnings often produce stormy weather as the heated air mass at the exterior rises above the cool air mass, creating clouds and storms.

4 0
3 years ago
What is the pressure of 20.5 mols of helium gas at 444 K and 999 L
Charra [1.4K]

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IDK

Explanation:

5 0
3 years ago
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [
sveta [45]

Explanation:

The given data is as follows.

     E^{o} = 0.483,     [Co^{3+}] = 0.173 M,

     [Co^{2+}] = 0.433 M,     [Cl^{-}] = 0.306 M,

     P_{Cl_{2}} = 9.0 atm

According to the ideal gas equation, PV = nRT

or,             P = \frac{n}{V}RT    

Also, we know that

                Density = \frac{mass}{volume}

So,         P = MRT

and,          M = \frac{P}{RT}

                    = \frac{9.0 atm}{0.0820 L atm/mol K \times 298 K}

                    = \frac{9.0}{24.436}

                    = 0.368 mol/L

Now, we will calculate the cell potential as follows.

          E = E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}

             = 0.483 - \frac{0.0591}{2} log \frac{(0.433)^{2}(0.368)}{(0.173)(0.306)^{2}}

             = 0.483 - 0.02955 log \frac{0.0689}{0.0162}

             = 0.483 - 0.02955 \times 0.628

             =  0.483 - 0.0185

             = 0.4645 V

Thus, we can conclude that the cell potential of given cell at 25^{o}C is 0.4645 V.

4 0
4 years ago
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