Answer:
100 g
Explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 400 g
Time (t) = 4 years
Half-life (t½) = 2 years
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Time (t) = 4 years
Half-life (t½) = 2 years
Number of half-lives (n) =?
n = t / t½
n = 4 / 2
n = 2
Thus, 2 half-lives has elapsed.
Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:
Original amount (N₀) = 400 g
Number of half-lives (n) = 2
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2² × 400
N = 1/4 × 400
N = 0.25 × 400
N = 100 g
Thus, the amount of the radioactive isotope remaing is the 100 g.
Answer:
9
Explanation:
The structure of fluorophore used in the experiments has been drawn in the attachment. And from the drawing counting we can say that there are 9 sp2-hybridized carbon atoms present. Fiuorophores are a fluorescent chemical compound that can re-emit light upon light excitation. Normally used to produce absorbance and emission spectra.
Explanation:
The glucose molecules changes as following:
C6H12O6 —> C2H5OH + CO2
There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:
C6H12O6 —> 2C2H5OH + CO2
There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:
C6H12O6 —> 2C2H5OH + 2CO2
Answer:
30 feet /second
Explanation:
60 feet/ 2 sec = 30 feet/sec
