Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Answer is (1) Produces H+ in aqueous solution
problem identification
hypothesis
experimentation
data collection and analysis
data interpretation
data presentation
conclusion
Answer:
1.73 atm
Explanation:
Given data:
Initial volume of helium = 5.00 L
Final volume of helium = 12.0 L
Final pressure = 0.720 atm
Initial pressure = ?
Solution:
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
P₁ × 5.00 L = 0.720 atm × 12.0 L
P₁ = 8.64 atm. L/5 L
P₁ = 1.73 atm
Answer:
It should be 1. 1.2 X 10^24
Explanation:
