The inducible isoform (NOS<span>-2) is calcium-independent and produces large amounts of </span>gas<span> that can be cytotoxic. </span>NOS<span> oxidizes the guanidine group of L-</span><span>arginine in a process that consumes </span>five<span> electrons and results in the formation of NO with stoichiometric formation of L-citrulline. </span>
Answer:
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)
Explanation:
All the following are oxidation–reduction reactions except:________
a. H₂(g) + F₂(g) → 2HF(g). Redox. H is oxidized and F is reduced.
b. Ca(s) + H₂(g) → CaH₂(s). Redox. Ca is oxidized and H is reduced.
c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g). Redox. K is oxidized and H is reduced.
d. 6Li(s) + N₂(g) → 2Li₃N(s). Redox. Li is oxidized and N is reduced.
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number
Number 1 do a data table on paper and do as it was said, then go into Microsoft Excel and put your chart in. For 2 make a line graph(as stated) and do what it tells you.For 3 do as said with the line graph or whatever it should look somewhat like a math chart(the coordinate plane) and do as said. For number 4 help him/her study the charts and analyze the difference between the two. And for number 5 make him/her choose one element and research and write down the questions.
Answer: 25 moles of product are formed.
Explanation:
The balanced chemical equation will be :
According to stoichiometry :
1 mole of 
require = 3 moles of 
Thus 5 moles of 
will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 4 moles of reactant give = 5 moles of product
Thus 20 moles of reactants will give =
of products
