Answer:
A)P5010
Explanation:
Penta- means 5 and deca/deco- means 10
Answer:
............................................
Explanation:
............................................
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
Answer:
A beaker
Step-by-step explanation:
Specifically, I would use a 250 mL graduated beaker.
A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.
You don't need precisely 100 mL solution.
If the beaker is graduated, you can easily measure 100 mL of the stock solution.
Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).
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
