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Rainbow [258]
3 years ago
11

Identify the characteristics of tornadoes and hurricanes

Chemistry
2 answers:
borishaifa [10]3 years ago
7 0

Tornado:

<u><em>-Process begins when the ground is heated to high temperatures.</em></u>

<u><em>-Common in the Plains states of the United States.</em></u>

<u><em>-Travels relatively short distances before losing its energy.</em></u>

Hurricane:

<u><em>-Develops over warm ocean water heated by the sun.</em></u>

<u><em>-Equivalent of a typhoon.</em></u>

<u><em>-Loses strength as it moves from water to land. </em></u>

Strong tornadoes usually occur over land while hurricanes almost always form over the ocean.

The Great Plains, the Midwest, the Mississippi Valley and the southern United States are all areas that are vulnerable to tornadoes.

The only difference between a hurricane, a cyclone, and a typhoon is the location where the storm occurs.

Hurricanes may lose strength over land because of cool temperatures and a lack of moisture.

Reil [10]3 years ago
4 0
Identify the characteristics of tornadoes and hurricanes.

1)develops over warm ocean water 
heated by the Sun<span>travels relatively short distances before </span>
losing its energyequivalent of a typhoon<span>loses strength as it moves </span>
from water to land<span>common in the Plains states of </span>
the United Statesprocess begins when the ground
<span>is heated to high temperatures</span>
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24 hours

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A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,
lara31 [8.8K]

Answer:

The answer to your question is:   T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L                    V2 = 2.78 L

P1 = 2.40 atm               P2 = 1.97 atm

T1 = 325°K                    T2 = ?

Formula

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Process

            T2 = (P2V2T1) / (P1V1)

            T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

            T2 = 1779.895 / 7.56

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4 0
3 years ago
Which are characteristics of metals that result from their bonding? Check all that apply.
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Conduct Electricity and Are malleable

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The freezing point of a nonelectrolyte solution containing 30.0 g of a solute dissolved in 250.0 g of water is observed to be -2
scZoUnD [109]

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Molar mass of solute is 89.28 g/m

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Colligative property of freezing point depression to solve this:

ΔT = Kf . m . i

i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.

m = molalilty (mol of solute/1kg of solvent

ΔT = T° freeze pure solvent - T° freeze solution

0°C - (-2.50°C) = 1.86 °C/m . m

2.50°C / 1.86 m/°C = m

1.34 mol solute/kg solvent = m

This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.

1000 g ____ 1.34 moles

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To find the molar mass, we divide mass / moles

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3 years ago
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