Answer:
6.9399amu
Explanation:
Isotopes are elements with same atomic number but different atomic masses.
To determine the average atomic mass one must multiply the absolute atomic mass (amu) of each isotope times its absolute atomic mass. This will give the weight average contribution of each isotope. The sum of the weight average contributions gives the final weight average atomic mass of the element. The wt. avg. atomic mass is the value posted on the periodic table.
It is recommended that a table be sketched in the following form ...
Isotope | % fractional amt. X Isotopic mass(amu) => Wt. Avg**.
Li-6 7.59% => 0.0759 x 6.015amu => 0.4565
Li-7 92.41% => 0.9241 x 7.016amu => 6.4835
*amu => atomic mass units ∑ Li-6 + Li-7 = 6.9399amu
**Wt.Avg. => weight average contribution of isotope = fractional amt x Isotopic mass(amu)
*** Final Wt. Avg. = ∑Wt.Avg. Contributions (value posted of periodic table)
Answer:
Q1. a) 4Al + 3O₂ ➟ 2Al₂O₃
b) 7.4 moles
c) 11.1 moles
Explanation:
To balance an equation, ensure that the total number of atoms of each element on both sides are equal.
Al + O₂ ➟ Al₂O₃
On the left side of the arrow, you would find the reactants while the product(s) is found on the left hand side.
<u>Reactants</u>
Al atoms: 1
O atoms: 2
<u>Product</u>
Al atoms: 2
O atoms: 3
After balancing,
4Al + 3O₂ ➟ 2Al₂O₃
We have 4 Al atoms and 6 O atoms on both sides.
b) The balanced equation tells us the mole ratio of Al to Al₂O₃.
Al: Al₂O₃
= 4: 2 (÷2 throughout)
= 2: 1
This means that for every 1 mole of Al₂O₃, 2 moles of Al is needed.
Since we need 3.7 moles of Al₂O₃,
number of moles of Al needed
= 2×3.7
= 7.4
c) 4Al + 3O₂ ➟ 2Al₂O₃
For every 4 moles of Al, 3 moles of O are needed.
For each mole of Al,
number of moles of O needed
= 3÷4
= 0.75
For 14.8 moles of Al,
number of moles of O required
= 0.75 ×14.8
= 11.1
In setting up your equipment in front of the fume hood to eliminate
any vapors during the experiment. Use an iron ring stand with iron ring to back
up the clay triangle to clasp the crucible and lid. Alter the height of the
iron ring so the end of the crucible is about one centimeter above the inner
blue cone of the strong flame.
Answer:
c it has the qualities of all the above