Answer:
The amount of delta-glucono lactone present after the first 1 hour = 54.88g
Step-by-step explanation:
dy/dt = -0.6y where y₀ (corresponding to t=0) is 100g
dy/dt = -0.6y
dy/y = -0.6dt
Integrating the left hand side from y₀ to y and the right hand side from 0 to t.
In (y/y₀) = -0.6t
(y/y₀) = e⁻⁰•⁶ᵗ
y = y₀ e⁻⁰•⁶ᵗ
But y₀ = 100g
y = 100e⁻⁰•⁶ᵗ
At t = 1,
y = 100e⁻⁰•⁶ = 54.88g
The amount of delta-glucono lactone present after the first 1 hour = 54.88g