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Ket [755]
2 years ago
5

In a bowl of 50 skittles, you have 12 yellow, 9 orange, 11 red, and 13 green. The rest of the skittles are purple. What percenta

ge of the skittles are red? What percentage of the skittles are not green?
Chemistry
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

percentage of red skittles = 22%

percentage of skittles that are not green = 74%

Explanation:

total skittles = 50

yellow = 12

orange = 9

red = 11

green = 13

purple = 50-12-9-11-13

= 5

1) percentage of red skittles

\frac{red \: skittles}{total \: skittles}  \times 100

\frac{11}{50}  \times 100

\frac{22}{10}

22\%

2) percentage of skittles that are not green

\frac{total \: skittles - green \: skittles}{total \: skittles} \times 100

\frac{50 -13 }{50}  \times 100

\frac{37}{50}  \times 100

74\%

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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
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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.

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