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klasskru [66]
3 years ago
10

Which is the term for how vegetation influences precipitation?

Chemistry
1 answer:
Scilla [17]3 years ago
3 0
The answer to this question is A- evaporation
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Which type of microscope can be used to view cellular organelles such as the endoplasmic reticulum and Golgi?
Natalija [7]
Transmission electron microscope.
3 0
3 years ago
PLS HELP THIS IMPORTANT
Ksenya-84 [330]

Answer:

Explanation:

Physical change is a temporary and reversible change in which the physical properties of the substance changes without altering the composition of the substance Eg Melting of ice while chemical change is a permanent and not so easily reversible change in which the identity of the substance is altered to produce one or more substances Eg Combustion

8 0
2 years ago
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what is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that the density of mercury is 13.6 g/ml?
lilavasa [31]
24 square root . i know im wrong but so what 
6 0
3 years ago
A sample of 19 mg of 9-fluorenone is dissolved in 5 mL of diethyl ether. The resulting organic
maks197457 [2]

Answer:

15.70mg would remain

Explanation:

Partition coefficient is used to extract or purify a solute from a solvent selectively to avoid interference from other substances. For the problem, formula is:

Kp = Concentration 9-fluorenone in ether / Concentration of solute in H₂O

After the solute, 9-fluorenone, is extracted with water, the mass that remains in ether is:

(19mg - X)

<em>Where X is the mass that now is in the aqueous phase</em>

Replacing in Kp formula:

9.5 = (19mg - X) / 5mL / (X /10mL)

0.95X = 19mg - X / 5mL

4.75X = 19 - X

5.75X = 19

X = 19 / 5.75

X = 3.30mg

That means 9-fluorenone that remain in the ether layer is:

19mg - 3.30mg =

<h3>15.70mg would remain</h3>
8 0
3 years ago
Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest
balu736 [363]

Answer:

Cost to supply enough vanillin is 3.2\$

Explanation:

Threshold limit of vanillin in air is 2.0\times 10^{-11}g per litre means there should be 2.0\times 10^{-11}g of vanillin in 1L of air to detect aroma of vanillin.

1ft^{3}=28.32L

So, 5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L

So amount of vanillin should be present to detect = (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g

As cost of 50 g vanillin is 112\$ therefore cost of  (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)gvanillin = (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$

5 0
3 years ago
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