Answer:
more electron deficient
Explanation:
The nitro group is an electron withdrawing group. It withdraws electrons from the pyridine ring by resonance.
This electron withdrawal by resonance makes the pyridine ring less electron rich or more electron deficient.
Hence, the nitro group makes the pyrinde ring more electron deficient
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
<span>The elements on the Periodic Table are arranged in order of increasing "Atomic number"
In short, Your Answer would be Option B
Hope this helps!</span>
6 sodium and 6 Bromine in 6NaBr
