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:
number of protons and neutrons
Explanation:
Answer:
Solving by the method of exponential growth.
bacteria = 2
after one hr = 2² = 4
after 2nd hr = 2³ = 8
after 3rd hr = 2⁴ = 16
after 4th hr = 2⁵ = 32
<span>Soil is partially the result of the physical and chemical weathering of its parent rock. The minerals found in the soil were either in that parent rock, or they were formed from the weathering products of the parent rock.</span>
An acid is deemed strong if it can readily or easy "donate" a proton (H+) to the other ions in the solutions. Also, to donate or lose the proton or H+, the acid must dissociate (split into ions) in the solution. The more it can readily dissociate, the stronger the acid is.