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3 years ago

Moving water carrying away small

Chemistry

2 answers:

Shtirlitz [24]3 years ago

8 0

Answer:

Your answer should be erosion.

Explanation:

Erosion is caused mainly by water. Moving water picks up tiny fragments of rock and soil and carries them away. Wind and glaciers can also erode sediments. Erosion shapes Earth's surface.

kondaur [170]3 years ago

7 0

Answer:

erosion:)

Explanation:

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k0ka [10]

Answer:

I think it's C or D Sorry if I'm Wrong

Explanation:

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3 years ago

Pls big test What are openings in the ground that release energy from deep inside the planet? * 1clouds 2volcanoes 3sun 4hurrica

9966 [12]

Answer:

2. Volcanoes

Explanation:

The interior of some planets (e.g earth) contains some hot flowing fluids called lava which has a very high thermal energy. The lava as a means of escaping the confinement within the planet, burst out by creating an opening through the ground to the surface of the planet to release the energy (lava).

This process in which the energy is released to the surface of the planet is termed volcanic eruptions. Which occurs majorly in some susceptible regions of the earth.

The openings in the ground that release energy from deep inside the planet is thus called volcanoes.

8 0

3 years ago

Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy

gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure , Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$ ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

$Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

$2P=PE^{6.4 \times 10^{-4}P}$

$2=E^{6.4 \times 10^{-4}P}$

$\ln 2 = 6.4 \times 10^{-4}P$

∴ P = 1083.04 atm

c). dG = Vdp -S dt at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp$

$=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

$=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol

7 0

3 years ago

If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?

Darina [25.2K]

The question is incomplete, here is the complete question:

A bottle of rubbing alcohol having aqueous solution of alcohol contains 70% (v/v) alcohol. If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?

<u>Answer:</u> The amount of water present in the given bottle of rubbing alcohol is 283.8 mL

<u>Explanation:</u>

We are given:

Volume of bottle of rubbing alcohol = 946 mL

70% (v/v) alcohol solution

This means that 70 mL of rubbing alcohol is present in 100 mL of solution

Amount of water present in solution = [100 - 70] = 30 mL

Applying unitary method:

In 100 mL of solution, the amount of water present is 30 mL

So, in 946 mL of solution, the amount of water present will be = $\frac{30}{100}\times 946=283.8mL$

Hence, the amount of water present in the given bottle of rubbing alcohol is 283.8 mL

4 0

3 years ago

Which characteristic is a chemical property?

Mrac [35]

I am answering A question

8 0

3 years ago

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