This question is a dilution problem. In order to prepare a certain concentration and volume from a stock solution, you have to use the equation,
M1V1 = M2V2
From the equation, you need to calculate the volume you need from the stock solution to be able to have the diluted concentration then dilute it to a certain volume.
(1.0 M) (V1) = (0.1 M) (100 mL)
V1 = 0.1 mL is needed from the 1.0 M HCl solution
Since the measurement is not changing, the answer is 100 mL. Hope this helps.
40 g Ar.......6.023*10^23 atoms
x g Ar.......3.8*10^24 atoms
x=40*3.8*10^24/(6.023*10^23)=252,365g Ar
The correct answer is 12.2% BaO.
The solution is found by dividing the mass of the BaO, which is 25.8 grams, by the total mass of the solution, which is 212 grams, then multiplying it by 100 to get the percentage:
Answer:
Average atomic mass of chlorine is 35.48 amu.
Explanation:
Given data:
Percent abundance of Cl-35 = 76%
Percent abundance of Cl-37 = 24%
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 = (76×35)+(24×37) /100
Average atomic mass = 2660 + 888 / 100
Average atomic mass = 3548/ 100
Average atomic mass = 35.48 amu
Average atomic mass of chlorine is 35.48 amu.
