Answer:
162 kJ
Explanation:
The reaction given by the problem is:
- NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJ
If we turn it around, we have:
- 3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJ
If we think now of HOBr and NH₃ as our reactants, then now we need to find out <u>which one will be the </u><em><u>limiting reactant</u></em> when we have 9 moles of HOBr and 2 moles of NH₃:
- When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our <em>reactant in excess</em>, thus NH₃ is our limiting reactant.
-81 kJ is our energy change when there's one mol of NH₃ reacting, so we <u>multiply that value by two when there's two moles of NH₃ reacting</u>. The answer is 81*2 = 162 kJ.
Maximum number of water molcules which can take part in forming hydrogen bonding interaction with asparagine are 13. Following structure shows the interactions. Lone pair of electrons act as hydrogen bond acceptors and hydrogen atoms attached to heteroatoms acts as hydrogen bond donors.
The solution before dilution and after dilution contains same number of moles, and water is added for dilution.
Option B
<h3><u>Explanation:</u></h3>
Suppose before dilution, the solution contains x moles of KCl in Y liter of water. Now as the concentration got halved, then the solution contains x moles of KCl in 2Y kiters of solution. So the number of moles of KCl in the solution remained constant.
Again, as the solution is diluted to half of the concentration, water must have been added with the solution to make it dilute.
Answer:
D
Explanation:
D Is The Answer Fella, My Head Hurt Really Bad
Answer:
It is used as a catalyst.
