B. He would be considered an Organic Chemist since Organic Chemistry is the study of Carbon and its compounds.
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .139 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.139-x ≈ 0.139
4.5x10^-4 = x^2/0.139
Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.007M/0.139M = .0503 or
≈5.03% dissociation.
Answer:
The energy released will be -94.56 kJ or -94.6 kJ.
Explanation:
The molar mass of methane is 16g/mol
The given reaction is:
the enthalpy of reaction is given as ΔH = -890.0 kJ
This means that when one mole of methane undergoes combustion it gives this much of energy.
Now as given that the amount of methane combusted = 1.70g
The energy released will be:
Answer:
The answer is "Choice A and Choice B"
Explanation:
The Zero-Order reactions are usually found if a substrate, like a surface or even a catalyst, is penetrated also by reactants. Its success rate doesn't depend mostly on the amounts of the various reaction in this reaction.
Let the Rate = k
As 
doesn't depend on reaction rate, a higher reaction rate does not intensify the reaction.
By the rate 
the created based and the reaction rate is about the same.
