Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
Grams. It is a smaller unit.
The balanced chemical reaction is:
<span>N2 + 3H2 = 2NH3 </span>
We are given the amount of H2 being reacted. This will be our starting point.
26.3 g H2 (1 mol H2 / 2.02 g H2) 2 mol
O2/3 mol H2) ( 17.04 g NH3 / 1mol NH3) = 147.90 g O2
Percent yield = actual yield / theoretical
yield x 100
Percent yield = 79.0 g / 147.90 g x 100
Percent yield = 53.4%
You should go on yahoo answers and ask it! Someone will get back to you
