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hram777 [196]
3 years ago
5

You are a drop of water going through the water cycle. Where do you go? How do you get there? What processes do you go through?

Include at least three phases of the water cycle . I do know what the phases of the water cycle,so i don't need the phases of the water cycle. I just need a righting
Chemistry
2 answers:
katovenus [111]3 years ago
7 0

Answer:

Explanation:

Hi

enyata [817]3 years ago
6 0

→Answer:

evaporation, condensation, precipitation

Explanation:

this diagram may help

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A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of
ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = V_1=10.0 L

Initial pressure of the gas = P_1=4.0 atm

Final volume of the gas = V_2=5.0 L

Final pressure of the gas = P_2=8.0 atm

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

PV=k

The equation given by this law is:

P_1V_1=P_2V_2

P_1\propto \frac{1}{V_1}

P_1V_1=k_i

k_i=4.0 atm\times 10.0 L = 40.0 atm L

P_2\propto \frac{1}{V_2}

P_V_2=k_f

k_f=8.0 atm\times 5.0 L = 40.0 atm L

k_i=k_f=40.0 atm L

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

6 0
3 years ago
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
3 years ago
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