I think the answer you have selected is right, I'm not sure tho
Answer:
.079 moles of Nirogen gas (N2)
Explanation:
You can see from the equaton that each ONE mole of N2 produces TWO moles of NH3.
Find the number of moles of NH3 produced.
Using Periodic Table : Mole wt of NH3 = 17 gm/mole
2.7 gm / 17 gm/mole = .1588 moles
One half as many moles of N2 are needed = .079 moles
Thank you for posting your question here. Below is the solution:
HNO3 --> H+ + NO3-
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>
<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>
<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
