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3 years ago

Give me some examples of why is it important to understand how ocean currents flow.

Chemistry

1 answer:

IceJOKER [234]3 years ago

8 0

Answer:

By moving heat from the equator toward the poles, ocean currents play an important role in controlling the climate. Ocean currents are also critically important to sea life. They carry nutrients and food to organisms that live permanently attached in one place, and carry reproductive cells and ocean life to new places.

Explanation:

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List two general properties of molecular compounds

Katen [24]

Low melting points and boiling points. ...Low enthalpies of fusion and vaporization These properties are usually one or two orders of magnitude smaller than they are for ionic compounds.Soft or brittle solid forms. ...Poor electrical and thermal conductivity.

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3 years ago

What does B represent?

Mashutka [201]

Answer: Botanical Acid

Explanation:

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3 years ago

What did Ernest Rutherford discover? How was his model different from Thomson’s?

s2008m [1.1K]

Answer:

thomson developed the chocolate chip method which was the identification of the electrons in the core of an atom. Rutherford discovered that the core was only positive and that the electrons were floating outside of the core.

Explanation:

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3 years ago

(2k-1)²<br> what the answer pls.

Sedaia [141]

Answer:

4k^2-4k+1

Explanation:

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3 years ago

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, $(5.45*V_{1} ) = (820.7*1.41)$

$V_{1}=\frac{820.7*1.41}{5.45}= 212.33 ml$

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0

3 years ago