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alexdok [17]
3 years ago
7

The domain Archaea are unicellular prokaryotes and can be autotrophs or heterotrophs

Chemistry
1 answer:
nexus9112 [7]3 years ago
7 0

Answer:

I think its true I dont really know

Explanation:

true

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For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided
SVETLANKA909090 [29]

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

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How does the process of waste removal in a unicellular organism compare to the waste removal in a multicellular organism
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The digestive system of multicellular is made up of several organs that work together to break down food so that it may be used in the body. Unicellular organisms have organelles, similar to organs, which help to digest food that will be used by the cell
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At a certain temperature, the solubility of N2 gas in water at 4.07 atm is 95.7 mg of N2 gas/100 g water . Calculate the solubil
sertanlavr [38]

Answer: Thus the solubility of N_2 gas in water, at the same temperature, if the partial pressure of gas is 10.0 atm is 235mg/100g.

Explanation:-

The Solubility of N_{2} in water can be calculated by Henry’s Law. Henry’s law gives the relation between gas pressure and the concentration of dissolved gas.

Formula of Henry’s law,  C=k_{H}P.

k_{H}= Henry’s law constant = ?

The partial pressure (P) of N_{2} in water = 4.07 atm

\C= k_{H}\times P\\95.7mg=k_{H}\times 4.07

k_{H}=23.5

At pressure of 10.0 atm

C= k_{H}\times P\\C=23.5\times 10.0=235mg/100mg

Thus the solubility of N_2 gas in water, at the same temperature, is 235mg/100g

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