Reduction. B is the correct answer
Answer:
NH₄Cl (aq) + NaOH (aq) -> NaCl (aq) + H₂O (l) + NH₃ (g)
(NH₄)₂CO₃ (aq) + 2KOH (aq) -> K₂CO₃ (aq) + 2H₂O (l) + 2NH₃(g)
NH₄NO₃ (aq) + LiOH (aq) -> LiNO₃ (aq) + H₂O (l) + NH₃ (g)
Explanation:
The gas liberated when an alkali reacts with an ammonium salt is NH₃ (ammonia), not CO₂.
3 o2 electrons are the strongest than the iron due to molecular stability
TLDR: the answer is C. 22,920 years.
Half-life describes the amount of time for a radioactive substance to decay to one-half of the original substance’s weight. So, if we had 100g of C-14, after 5,730 years, only 50g remain; after another 5,730 years, only 25g would remain, and so on.
In this problem, we are meant to assume that the original amount of C-14 was 64g, and that, through decay, it forms N-14. We can figure out how many half lives have passed by figuring out how much 4 is out of 64 by dividing 64 by two repeatedly. Each time, count a half life.
64 - 32 (1 HL) - 16 (2 HL) - 8 (3 HL) - 4 (4 HL)
In this problem, 4 half lives have passed. We can now multiply this by the time for one half life to find how many years have passed.
4 x 5,730 = 22,920 years
Approximately 22,920 years have passed since the drawing was created.