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3 years ago

5

as you move down the levels of classification, from domain to species, the number of organisms in each group increases? True or

false if false change the underlined word

2 answers:

Tanzania [10]3 years ago

8 0

The statement is true because there are many types of species in one domain

bija089 [108]3 years ago

4 0

True because you will grow in size as you get bigger

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A group of parasites called myxozoans have traditionally been considered simple multicellular protists. However, because these o

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A group of parasites called myxozoans have traditionally been considered simple multicellular protists. However, because these organisms possess stinging cells, some scientists consider them to be animals in the phylum that is named for such cells, the <u>cnidarians</u>.

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Cnidarians are animals with radial symmetry, that is, their bodies can be divided, by different planes, into equal parts. It includes hydras, jellyfish, anemones and marine corals. Its body is a kind of sac with an opening (the mouth) surrounded by a circle of tentacles. Some have a dominant jellyfish phase, others polyp, or both. They present defensive nematocysts with stinging action.

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3 years ago

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2 years ago

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Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

$C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}$

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

Best regards.

7 0

3 years ago