2, 8,6 because it has to be in a configuration of 2,8,8
<span>Their orbitals are completely filled</span>
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
Answer:
P₂ = 299.11 KPa
Explanation:
Given data:
Initial volume = 600 mL
Initial pressure = 70.00 KPa
Initial temperature = 20 °C (20 +273 = 293 K)
Final temperature = 40°C (40+273 = 313 K)
Final volume = 150.0 mL
Final pressure = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 70 KPa × 600 mL × 313 K / 293K ×150 mL
P₂ = 13146000 KPa .mL. K /43950 K.mL
P₂ = 299.11 KPa
