What helps me to balance equations is to list the elements i have on each side of the equation, and use tally marks to see what I have and don't have. Then when I'm done balancing, I tally again to make sure everything matches up.
On the left side, you have 1 Al, and 2 O. On the right side, 1 Al and 3 O.
In order for the equation to balance, you need to place a 2 in front of the AlO on the right side. This would make the Al have 2 atoms and the O have six. On the left side, you need to place a 2 in front of the Al and a 3 in front of the O, making it six. Left side: 2 Al's 6 O's. Right side: 2 Al's and 6 O's. Matches!
The molar mass is 294.1527 g/mol
Answer:
Average atomic mass = 17.5 amu.
Explanation:
Given data:
X-17 isotope = atomic mass17.2 amu, abundance:78.99%
X-18isotope = atomic mass 18.1 amu, abundance 10.00%
X-19isotope = atomic mass:19.1 amu, abundance: 11.01%
Average atomic mass of X = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100
Average atomic mass = 1358.628 + 181 +210.291 / 100
Average atomic mass = 1749.919 / 100
Average atomic mass = 17.5 amu.
36.49 gm using law of equivalent proportion
