Answer:
A is 4.5grams greater than A'.
Step-by-step explanation:
The initial amount of chemical, A₀ = 12g
In simple step by step terms,
If the chemical dissolves, the amount reduces by 0.5g/s.
In 12s, amount that has dissolved = 12 × 0.5 = 6g and the amount remaining, A = 12 - 6 = 6g.
If the metal dissolves, the amount reduces by half of its weight every 4 seconds. But 12 seconds has three 4 seconds.
In 4 seconds, 12 g of chemical reduces by half of 12 = 6g. And amount remaining, A' = 12 - 6 = 6g.
In another 4 seconds, 6g of chemical reduces by half of 6 = 3g. Amount of chemical remaining, A' = 6 - 3 = 3g
In the final 4 seconds, 3 g of chemical reduces by half of = 1.5g. Amount of chemical remaining finally, A' = 3 - 1.5 = 1.5g
difference in the amount left from the two cases = A - A' = 6 - 1.5 = 4.5g
If the rate of reaction analysis method to this question is required;
The two key pieces of information given in the question gives the rate of reaction in form of amount of reactant chemical remaining for the two cases.
If the chemical dissolves,
The amount of chemical left at any time is A.
The rate of reaction in g/s = -0.5g/s
dA/dt = -0.5
dA = -0.5dt
Integrating the left hand side from A₀ to A, and the right hand side from 0 to t
A - A₀ = -0.5t
A = A₀ - 0.5t
For the 2nd case if the metal plate is dissolving,
The amount of chemical left at any time is A'
Half-Life = 4 secs
k = (In 2)/(half life) = (In 2)/4 = 0.1733/s
dA'/dt = -0.1733A'
dA'/A' = -0.1733 dt
Integrating from A₀ to A' and from 0 to t
In (A'/A₀) = -0.1733t
A'/A₀ = e⁻⁰•¹⁷³³ᵗ
A' = A₀e⁻⁰•¹⁷³³ᵗ
After t = 12s,
A = A₀ - 0.5t = 12 - 0.5(12) = 12 - 6 = 6g
A' = A₀e⁻⁰•¹⁷³³ᵗ = 12e⁻⁰•¹⁷³³⁽¹²⁾ = 12 × 0.125 = 1.5g
The difference in mass = A - A' = 6 - 1.5 = 4.5g
Hope this helps!!!
