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
Activation Energy. All chemical reactions, even exothermic reactions, need a certain amount of energy to get started. This energy is called activation energy. For example, activation energy is needed to start a car.
<span>The correct answer is 3: 0.10 M K3PO4(aq). The higher the concentration (or molarity), the higher the boiling point. Thus, the solution with the most moles will have the highest boiling point. SO4, PO4, and NO3 are all polyatomic ions, so by definition, they only have one mole. In K3PO4, K3 has 3 moles and PO4 has 1 mole, meaning all together it has 4 moles (more than any of the other options).</span>
D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3
dim? im not so fluent in this but i did research yesterday
