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Gala2k [10]
4 years ago
7

The earth's geochemical cycles are the ___ and ___

Chemistry
1 answer:
Softa [21]4 years ago
7 0

Answer:

Nitrogen and Carbon

Explanation:

Nitogen cycle and carbon cycle are the two biogeochemical cycles

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Energy resources are required for which of the following ?
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Identify three additional ways scientists learn about the Earth's interior.
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The Equator is an imaginary line around the middle of the Earth. It is halfway between the North and South Poles, and divides the Earth into the Northern and Southern Hemispheres. The Equator is the line of 0 degrees latitude. Each parallel measures one degree north or south of the Equator, with 90 degrees north of the Equator and 90 degrees south of the Equator. The latitude of the North Pole is 90 degrees N, and the latitude of the South Pole is 90 degrees S hope this helps
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Observe the image of hydrogen chloride. List the properties you see.
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Explanation:

The properties of the hydrogen chloride:

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  • It is clear, colorless and transparent.
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4 years ago
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How could you turn steam into a liquid?
marusya05 [52]

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4 years ago
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A natural water with a flow of 3800 m3/d is to be treated with an alum dose of 60 mg/L. Determine the chemical feed rate for the
svet-max [94.6K]

Explanation:

First, we will calculate the feed rate of alum as follows.

   \frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}

                  = 228000 g/day

Converting this amount into g/min as follows.

     \frac{228000 g}{1 day} \times \frac{1 day}{1440 min}

          = 158 g/min

Now, the chemical equation will be as follows.

    Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O

 \frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}

      = 0.151 mmol mmol SO^{2-}_{4}/L

\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}

           = 15.15 mg CaCO_{3}/L

For precipitate:

Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O

  \frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}

     = 7.88 Al(OH)_{3}/L

  \frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}

          = 29.9 Al(OH)_{3}/day

3 0
4 years ago
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