Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
A solution is turning blue means, it is turning it's behavior to Basic from Acidic. Reagent D must be a Basic component so it is increasing the pH of the solution. As reaction does not depend on the phase of the component, it could be anything i.e., Solid, Liquid or gas.
Hope this helps!
It's B) electrons, is the right answer
Answer:
No. Of Moles of zinc = m/Ar
= 13/ 65.38 = 0.198 moles
From balanced equation, Mole ration between CuSO4 and Zn is 1 : 1
So only 0.198 moles of CuSO4 reacts, it is in excess
Mass = no of Moles X Mr
Mass = 0.198 X 159.5 = 31.59 grams
Volume = mass m denisty
Volume j 31.59 / 3.6 = 8.78 ml
Explanation:
i think this wrong
Explanation:
Standards.
Assessments.
Accountability.
Professional Development.
School Autonomy.
Parent Involvement.
Learning Readiness.
Technology.
