Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Explanation:
Moles of N2 = 84/28 = 3.0mol
Moles of H2 = 29/2 = 14.5mol
Hence amount of ammonia produced = 6.0 * 17 = 102g.
Answer:
The correct option is: (D) would function as both an acid and a base
Explanation:
A carbon skeleton bonded to a amino group as well as a carboxyl group, will behave as an acid in basic medium and base in acidic medium. This is because the carboxyl group present in the compound will release a proton in basic medium and the amino group will accept a proton in the acidic medium.
<u>Therefore, a carbon skeleton which is covalently bonded to a carboxyl and amino group will behave as both acid and base.</u>
The answer is B. the occurrence of huge events in Earth's natural history
The geologic time scale is a system of chronological dating that relates geological strata to time. It is used by geologists, paleontologists, and other Earth scientists to describe the timing and relationships of events that have occurred during Earth's history.
