3Si + 2N2 --> Si3N4 (as given)
n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol
Therefore, N2 is limiting and Si is in excess
The molar ratio of 2N2:Si3N4 is 2:1
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2)
m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g
= 8.066g (4 significant figures)
(hopefully it is right, but double check in case i did something wrong) :)
Answer:
The pH of a solution that has a [OH-] of 5.08x10^-5 M is 5
Explanation:
just took the test
onedg2020
11. D
12.C
13. For 13 i would say B but not so sure
14. A
