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ZanzabumX [31]
3 years ago
12

Like in a cat's eye, the human pupil changes shape in response to changes in

Chemistry
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

B.

Our pupils contract and expand depending on the amount of light at any given time in order to avoid being blinded and to see better in the darkness.

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Someone plz help i need to do this assignment to pass the class
Natalija [7]

Answer:

1)0.2

2)0.72

3)0.01

Explanation:

Formula is 1)454÷2270

2)0.6×1.2

3)8÷800

8 0
2 years ago
Elements in the same period of the periodic table exhibit similar physical and chemical properties.
Vladimir [108]

Answer:

same valency electrons

Explanation:

example g 1 elements

5 0
3 years ago
Select the correct answer. How does an atom become a positively charged ion? A. By gaining one or more electrons B. By gaining o
ser-zykov [4K]

C. By losing one or more electrons

6 0
3 years ago
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Write a hypothesis statement about the relationship between the number of mushrooms and the concentration of heavy metals in the
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Write a hypothesis statement about the relationship between the number of mushrooms and the concentration of heavy metals in the soil ...


8 0
2 years ago
5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf
mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and pK_{a} is as follows.

               pH = pK_{a} + log \frac{base}{acid}

where,     pH = 7.4 and pK_{a} = 7.21

As here, we can use the pK_{a} nearest to the desired pH.

So,      7.4 = 7.21 + log \frac{base}{acid}

             0.19 = log \frac{base}{acid}

            \frac{base}{acid} = 1.55

1 mM phosphate buffer means [HPO_{4}] + [H_{2}PO_{4}] = 1 mM

Therefore, the two equations will be as follows.

           \frac{HPO_{4}}{H_{2}PO_{4}} = 1.55 ............. (1)

  [HPO_{4}] + [H_{2}PO_{4}] = 1 mM ........... (2)        

Now, putting the value of [HPO_{4}] from equation (1) into equation (2) as follows.

             1.55[H_{2}PO_{4}] + [tex][H_{2}PO_{4}] = 1 mM

                        2.55 [H_{2}PO_{4}] = 1 mM

                             [H_{2}PO_{4}] = 0.392 mM

Putting the value of [H_{2}PO_{4}] in equation (1) we get the following.

                     0.392 mM + [HPO_{4}] = 1 mM

                          [HPO_{4}] = (1 - 0.392) mM

                              [HPO_{4}] = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0
3 years ago
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