You need to prepare 2.00 l of 0.100 m na2co3 solution. the best procedure is to weigh out:

a. 21.2 g na2co3 and add 2.00 l of water to it.
b. 10.6 g na2co3 and add 2.00 l of water to it.
c. 21.2 g na2co3 and add water until the final solution has a volume of 2.00 l.
d. 10.6 g na2co3 and add water until the final solution has a volume of 2.00 l.

The answer to your question is letter C

Explanation:

Data

Volume = 2 L

Molarity = 0.100 M

Molecular weight Na₂CO₃ = (2 x 23) + (1 x 12) + (3 x 16)

                                           = 46 + 12 + 48

                                           = 106 g

Process

1.- Calculate the grams of Na₂CO₃ needed

                         106 g  1 mol

                           x        0.1 moles

                           x = (0.1 x 106) / 1

                           x = 10.6 g

2.- Calculate the grams of Na₂CO₃ needed for 2 liters of solution

                           10.6 g 1 liter

                            x          2 liters

                           x = (10.6 x 2) / 1

                           x = 21.2 grams of Na₂CO₃                        

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percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally