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Ugo [173]
3 years ago
7

I’ll give brainliest!

Biology
2 answers:
Papessa [141]3 years ago
8 0

Answer

Excess admissions into the air.

Dima020 [189]3 years ago
5 0

Answer:

Some causes of climate change and the timescale for it is global warming which is caused by green house gases, and pollution, and the timescale for it is prob 12 years ago, and the greenhouse effect causes are methane and carbon dioxide with toxins mixed up, and it lasts to 300 to 1000 years :3

Explanation:

:3

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Flura [38]
I think it's option c
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19.
-Dominant- [34]

Answer:

D, they are all the same size.

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Why doesn't our hands swell up due to osmosis , when we keep it in water for a long time ??
DerKrebs [107]
A cell wall is a permeable layer found in plant cells which acts as a membrane through which water molecules can pass through while animal cells do not have cell walls.. they just have a cell membrane which is selectively permeable to some chemicals or substances.. water does not move from the region of higher to lower concentration in animal cells

In short, plant cells have cell walls that allow osmosis while animal cells do not have cell walls for osmosis to take place

hope it helps you :)

Instead in our hands... something else happens rills or groove shaped lines are formed in our fingers.. it is an adaptation for our body to provide for some friction in water...

8 0
3 years ago
which of the following statements about of matter is true? a. a gas has a definite volume but not a definite shape. b. a liquid
Sever21 [200]

I believe the answer is B.

6 0
3 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}

Best regards.

7 0
3 years ago
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