You are given a test tube containing 10 mL of a solution with 8.4 × 107 cells/mL. You are to produce a solution that contains less than 100 cells/mL. What dilutions must you perform in order to arrive at the desired result?

(1). 1 ml of original solution to 99 ml of water = 8.4 ×

(2). 1 ml of second solution to 99 ml of water = 8.4 ×

(3). 1 ml of third solution to 99 ml of water = 84  

Explanation:

Volume of solution = 10 ml

We should perform the following dilutions.

(1). 1 ml of original solution to 99 ml of water = 8.4 ×

(2). 1 ml of second solution to 99 ml of water = 8.4 ×

(3). 1 ml of third solution to 99 ml of water = 84  

These are the three dilutions we have to perform.

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