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3 years ago

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

Read 2 more answers

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 $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g$ vanillin = $(2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$$

5 0

3 years ago