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3 years ago

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1. The part of plants that contains structures

2 answers:

Nonamiya [84]3 years ago

8 0

Answer No 1:

<u>Chloroplasts in the leaves</u> are The part of plants that contains structures

needed to absorb sunlight and carbon dioxide.

Answer No 2:

The <u>Chloroplast </u>cells inside this structure contain many green

organelles called<u> chlorophyll</u>.

Answer No 3:

Inside each of these organelles are stacks of <u>thylakoid </u> that

contain the green pigment.

Answer No 4:

<u>Chlorophyll</u> absorbs sunlight needed to power the light dependent reactions of photosynthesis

Explanation:

The process of photosynthesis can be described as a series of reactions in which carbon dioxide and water are converted in to food i.e glucose in the plants. Oxygen is the by product of this reaction. Certain reactions of this process occur in the presence of light and are termed as light dependent reactions. The reactions which do not require light are termed as light independent reactions.

The process of photosynthesis occurs in the leaves. Leaves have structures called chloroplast where photosynthesis occurs. The chlorophyll are present inside the chloroplast which rap the sun's light energy.

Lady_Fox [76]3 years ago

5 0

Answer:

1. leaf

2. chloroplasts

3. thylakoids

4. chlorophyll

Explanation:

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

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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.

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