How many particles are in 9.22 x 10^–37 mol of iron(III) oxide (Fe2O3)?
A. 5.54 x 10^12 mol Fe2O3
B. 1.60 x 10^23 mol Fe2O3
C. 1.47 x 10^–23 mol Fe2O3
D. 5.55 x 10^–13 mol Fe2O3
I think it is C?
5 answers
I am wondering if you typed the question right. You are given mole iron(III)oxide, and the answers are in mols? Makes no sense to me. And, if you have an moles iron oxide, the particles has to be greater than one. When counting, you have to start at one. Answer C is less than one.
I definitely typed the question right bc i pasted it but I wouldn't be surprised if I got the answer wrong... If it can't be less than one then it is between A and b right? so B?
6.022 * 10^23 particles per mol. Like period, Avagadro's number.
9.22 x 10^–37 mol Is not even a single particle, nonsense
Maybe 9.22 Mols????
9.22 mols * 6.022*10^23 particles/mol
= 55.5 * 10^23 = 5.55 * 10^24
As bobpursley pointed out, the question makes no sense.
9.22 x 10^–37 mol Is not even a single particle, nonsense
Maybe 9.22 Mols????
9.22 mols * 6.022*10^23 particles/mol
= 55.5 * 10^23 = 5.55 * 10^24
As bobpursley pointed out, the question makes no sense.
so which one should i choose lol
Here is how you work the problem and the answer the author THINKS you should get is D,
9.22E-37 x 6.02E23 = 5.55E-13; however, as everyone above has pointed out to you that can't be right. Your prof goofed but if I told him/her I would be very diplomatic about it.
9.22E-37 x 6.02E23 = 5.55E-13; however, as everyone above has pointed out to you that can't be right. Your prof goofed but if I told him/her I would be very diplomatic about it.
1 answer