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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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Trava [24]

Answer: 2.52 M

Explanation:

The product of molarity (moles/litre) and volume in litres yields moles, and the numbers of moles in two solutions means dilute and concentrated are equal, which is expressed by the following equation: 

M_1V_1=M_2V_2

$$Given that\\M $1=12.0 \mathrm{M}$ or mole $/ \mathrm{L}$\\$\mathrm{V} 1=420 \mathrm{ml}$\\$\mathrm{M} 2=$ ?\\$\mathrm{V} 2=2.0 \mathrm{~L}$ or $2000 \mathrm{ml}$\\\\$\mathrm{M} 1 \mathrm{~V} 1=\mathrm{M} 2 \mathrm{~V} 2$\\$\mathrm{M} 2=\mathrm{M} 1 \mathrm{~V} 1 / \mathrm{V} 2$\\$=12.0 * 420 / 2000$\\$=2.52 \ \mathrm{M}$

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2 years ago
How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
ankoles [38]
The chemical reaction equation for this is 

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6, 

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x \frac{1 mole XeF6}{3 moles H2} x \frac{245.28 g XeF6}{1 mole XeF6} = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
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3 years ago
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