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.
Answer:
209.68
Explanation:
The only number that is relevant (though the rest are quite interesting) is the last one 1.98 * 10^24
1 mole of Barium Acetate Contains 6.02*10^23 particles.
There are 4 moles of carbon to every mole of Barium Acetate.
1.98 * 10^24 atoms / (4*6.02*10^23)
0.8223 moles of Ba(C2H3O2)2
Ba = 137
4C = 4*12 48
6H = 6*1 6
4O = 4*16 64
1mole 255 grams
0.8223 * 255 = 209.68 grams
I have used rounded masses for these elements depending on the periodic table you use. Go through the question with your masses to get a more accurate answer. My answer will not differ by much. It is a guide.
Atomic number:)!!!!!!!!!!!!!!!
Answer:
Protons=7, Neutrons=6, Electrons=6
Explanation:
Hope this helps!!
