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Sever21 [200]
2 years ago
9

Describe the role of oceans in the water cycle?

Chemistry
2 answers:
Ainat [17]2 years ago
4 0

Answer: The ocean plays a key role in this vital cycle of water. The ocean holds 97% of the total water on the planet; 78% of global precipitation occurs over the ocean, and it is the source of 86% of global evaporation. ... Water evaporates from the surface of the ocean, mostly in warm, cloud-free subtropical seas.

Explanation:

Hope this helps!

marusya05 [52]2 years ago
4 0
They are basically a heat sink for our planet, will soon to be a place for commercial grade mining, literally absorbs most of our carbon.
Those are some of the thing I can just think of from my aice marine class I’m sure there’s more I hope this helps.
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Which system consists of nerves that initiate communication?
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Magicians often use ‘flash powder’ in their acts. Flash powder contains 69.0% potassium chlorate (31.9% K, 28.9% Cl, 39.2% O) an
mafiozo [28]

Answer:

\rm K: approximately 22.0\%.

\rm Cl: approximately 19.9\%.

\rm O: approximately 27.0\%.

\rm Al: 31.0\%.

Explanation:

Consider a 100\; \rm g sample. There would be 69.0\% \times 100\; \rm g = 69.0\; \rm g of \rm KClO_{3} and 100.0\; \rm g - 69.0\; \rm g = 31.0\; \rm g of \rm Al.

Out of that 69.0\; \rm g of \rm KClO_{3}:

31.9\% (by mass) would be \rm K: 31.9\% \times 69.0\; \rm g \approx 22.0\; \rm g.

28.9\% (by mass) would be \rm Cl: 28.9\% \times 69.0\; \rm g \approx 19.9\; \rm g.

39.2\% (by mass) would be \rm O: 39.2\% \times 69.0\; \rm g \approx 27.0\; \rm g.

Overall, the composition (by mass) of each element in this mixture would be:

\rm K:

\begin{aligned} & \frac{31.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 31.9\% \times 69.0\% \\ \approx \; & 22.0\%\end{aligned}.

\rm Cl:

\begin{aligned} & \frac{28.9\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 28.9\% \times 69.0\% \\ \approx \; & 19.9\%\end{aligned}.

\rm O:

\begin{aligned} & \frac{39.2\% \times 69.0\; \rm g}{100\; \rm g} \\ =\; & 39.2\% \times 69.0\% \\ \approx \; & 27.0\%\end{aligned}.

\rm Al:

\begin{aligned}& \frac{100\; \rm g - 69.0\; \rm g}{100\; \rm g} \\ =\; & 100\% - 69.0\% \\ =\; & 31.0\%\end{aligned}.

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