manganese The metal that is used as a sacrificial electrode to prevent the rusting of iron is manganese.
<span>they both had their conclusions based on solid evidence</span>
Hi there!
p = e-3
s = f-1
f = i-7
d = g-5
Hope that helps!
Brady
Answer:
First
divide each element by its Molecular Mass to get their respective moles
Then Divide through by the lowest of the moles
You'll have the ratio of Carbon Hydrogen and Oxygen to be
C2H3O
Given Molecular Mass=184.27
C2H3On=184.27
n(12x2 + 1x3 + 16) =184.27
Evaluating this... You'll have n=4.3
Pls check if you assigned the correct value to each element
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