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2 years ago

14

Plants are eaten by grasshoppers, which are eaten by mice, which are eaten by snakes.

2 answers:

MrMuchimi2 years ago

8 0

If the Grasshoppers go missing, then the mice will lose their food source and eventually die. Which means the snakes will also lose their food source, which is mice.

joja [24]2 years ago

4 0

Answer: The population of mice and snakes will also reduce and plants population will increase.

Explanation:

Plants are producers that synthesize food by the help of the process known as photosynthesis. This process provides food to grass hoppers.

Grasshoppers are eaten by mice and mice is eaten by snakes. If the grasshoppers will be removed from the ecosystem then rest all the other organisms will be decreased in number.

Plants will increase in number as there will be no grasshopper to feed on them.

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2 years ago

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.

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

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2 years ago

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wlad13 [49]

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