A sample of rubidium contains two different isotopes. 72.15% of the sample

Chemistry

1 answer:

Leona [35]2 years ago

4 0

Answer:

Average atomic mass = 85.557 amu.

Explanation:

Given data:

Percent abundance of Rb-85 = 72.15%

Percent abundance of Rb-87 = 27.85%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (72.15×85)+(27.85×87) /100

Average atomic mass = 6132.75 + 2422.95 / 100

Average atomic mass = 8555.7 / 100

Average atomic mass = 85.557 amu.

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Calculate the Ph and the POH of an aqueous solution that is 0. 040 m in HCl(aq) and 0. 075 m in HBr(aq) at 25 °C.

LenaWriter [7]

pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.

pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.

pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.

Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.

Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :

HCL(0.040M)----> H+(0.040M) +CL-(0.040M)

HBr(0.075M)----> H+(0.075M) +Br-(0.075M)

so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.

pH is calculated as:

pH = -log[H+]

Substituting values in the equation:

log(0.115M)= 0.939 pH

pOH is calculated as:

14 - pH = pOH

Substituting values in the equation above:

14 - 0.939= 13.061 pOH