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Softa [21]
3 years ago
11

I really need someone to explain how to find the average atomic mass.... the gizmos was really confusing because they also had w

eighted mass.

Chemistry
1 answer:
evablogger [386]3 years ago
5 0

Answer:

Option C. 52.057

Explanation:

The following data were obtained from the question:

Isotope >> Mass number > Percentage

A (Cr-50) > 50 >>>>>>>>>> 4.3

B (Cr-52) > 52 >>>>>>>>>> 83.8

C (Cr-53) > 53 >>>>>>>>>> 9.5

D (Cr-54) > 54 >>>>>>>>>> 2.4

Average atomic mass =?

The average atomic mass of chromium, Cr can be obtained as follow:

Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]

Atomic mass of Cr = [50×4.3)/100] + [52×83.8)/100] + [53×9.5)/100] + [54×2.4)/100]

= 2.15 + 43.576 + 5.035 + 1.296

Atomic mass of Cr = 52.057

Therefore, the atomic mass of chromium, Cr is 52.057

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Calculate the pH after 10 mL of 1.0 M sodium hydroxide is added to 60 mL of 0.5M acetic acid.
baherus [9]

Explanation:

It is given that the total volume is (10 mL + 60 mL) = 70 mL.

Also, it is known that M_{1}V_{1} = M_{2}V_{2}

Where,    V_{1} = total volume

               V_{2} = initial volume

Therefore, new concentration of CH_{3}COOH = \frac{M_{2}V_{2}}{V_{1}}

                                        = \frac{60 \times 0.5}{70}

                                        = 0.43 M

New concentration of NaOH = \frac{M_{2}V_{2}}{V_{1}}

                                               = \frac{10 \times 1.0}{70}

                                               = 0.14 M

So, the given reaction will be as follows.

              CH_{3}COOH + OH^{-} \rightarrow CH_{3}COO^{-} + H_{2}O

Initial:             0.43          0.14                     0

Change:          -0.14        -0.14                    0.14

Equilibrium:    0.29          0                       0.14

As it is known that value of pK_{a} = 4.74

Therefore, according to Henderson-Hasselbalch equation calculate the pH as follows.

           pH = pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

                 = 4.74 + log \frac{0.14}{0.29}

                 = 4.74 + (-0.316)

                 = 4.42

Therefore, we can conclude that the pH of given reaction is 4.42.

6 0
3 years ago
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