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2 years ago

Does 4 have an infinate number of significant digits?

Chemistry

1 answer:

lisabon 2012 [21]2 years ago

6 0

4 is Significant! It doesn’t have 0’s

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A certain food web is made up of turtles, pond weeds, water fleas, and tadpoles. Which organism is most likely the producer in t

Simora [160]

The pond weeds are the producers because they are a plant

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2 years ago

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G what is the difference between the sidechains of leucine and isoleucine? select one:

statuscvo [17]

c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.

The structure of leucine is CH3CH(CH3)CH2CH(NH2)COOH.

The structure of isoleucine is CH3CH2CH(CH3)CH(NH2)COOH.

In leucine, the CH3 group is two carbons away from the α carbon; in isoleucine, the CH3 group is on the carbon next to the α carbon.

Thus, isoleucine has the closer branched carbon.

“One is charged, the other is not” is incorrect. Both compounds are uncharged.

“One has more H-bond acceptors than the other” is incorrect. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.

“They have different numbers of carbon atoms” is incorrect. They each contain six carbon atoms.

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3 years ago

What is empirical formula for CLO2

Natasha2012 [34]

Answer:

chlorine dioxide is empirical formula for CLO2.

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2 years ago

Which of the following organisms would NOT be in the first trophic level of an energy pyramid?

faust18 [17]

dog!

this is because the first energy comes from plants bc they can make their own food— therefore, energy :)

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3 years ago

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How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut

san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = 8 = Concentration in Ethyl acetate / Concentration in water

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

Where X is the amount of material in grams that comes to the organic phase.

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

In the second extraction, you will extract 0.50g of material

Thus, after the two extraction you will lost:

4.44g + 0.50g = 4.94g of material

6 0

2 years ago