According to this formula:
(P1V1) / T1 = (P2V2) / T2
convert T from C° to Kelvin:
T1 = 83 + 273 = 356 K
T2= 96 + 273 = 369 K
convert P from torr to atm:
1 torr = 0.00131578947 atm<span>
p1 = 0.839474 atm
P2 = 1.415789 atm
By substitution in the previous formula:
(0.839474 x 10.6 ) / (356) = ( 1.415789 x V2 ) / 369
So:
V2 = 6.5 L</span>
Noble gas notation for molybdenum:
[Kr] 4d^5 5s^1
Answer:
d. 12.3 grams of Al2O3
Explanation:
Based on the reaction:
4Al + 3O2 → 2Al2O3
<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>
<em />
To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:
<em>Moles Al -Molar mass: 26.9815g/mol-</em>
6.50g * (1mol / 26.9815g) = 0.241 moles Al
<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>
0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3
0.120 moles Al2O3 * (101.96g / mol) =
12.3g of Al2O3 are produced.
Right answer is:
<h3>d. 12.3 grams of Al2O3
</h3>
A) Head to tail joining of monomers. :) (confirmed correct answer, I took the test)
