Answer: 8.28g Na
Explanation: use ideal gas law
PV= nRT
Solve for moles of Cl2
n= PV/ RT
Substitute:
= 1 atm x 4.0 L / 0.08205 L.atm/ mol. K x 273 K
= 0.18 moles Cl2
Do stoichiometry to solve for m of Na
2 Na + Cl2 => 2 NaCl2
=0.18 moles Cl2 x 2 mol Na/ 1 mol Cl2 x 23g Na / 1 mol Na
= 8.28 g Na.
Answer : The percent abundance of the heaviest isotope is, 78 %
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
As we are given that,
Average atomic mass = 48.68 amu
Mass of heaviest-weight isotope = 49.00 amu
Let the percentage abundance of heaviest-weight isotope = x %
Fractional abundance of heaviest-weight isotope =
Mass of lightest-weight isotope = 47.00 amu
Percentage abundance of lightest-weight isotope = 10 %
Fractional abundance of lightest-weight isotope =
Mass of middle-weight isotope = 48.00 amu
Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %
Fractional abundance of middle-weight isotope =
Now put all the given values in above formula, we get:
Therefore, the percent abundance of the heaviest isotope is, 78 %