The half life for radioactive can be calculated as:
N /N0 = (1 /2) ^ n
n = T /T half
According to question there are n number of half life are present which would result in remaining amount of element as n.
Answer:
See explanation
Explanation:
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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
Answer:
(4) option is correct
Explanation:
Given that,
The conformation of 2-methyl-3-pentanol, as viewed along the C₂-C₃ bond.
We need to find the Newman structures
Using given structures
The structure of 2- methyl-3- pentanol is shown in figure.
Through C₂-C₃ carbon shown the structure of Newman projection in figure.
Here, carbon 2 is in front side and carbon 3 is in back side in the figure.
Hence, (4) option is correct.
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