The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
Answer:
44.7 kWh
Explanation:
Let's consider the reduction of Al₂O₃ to Al in the Bayer process.
6 e⁻ + 3 H₂O + Al₂O₃ → 2 Al + 6 OH⁻
We can establish the following relations:
- The molar mass of Al is 26.98 g/mol.
- 2 moles of Al are produced when 6 moles of e⁻ circulate.
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant).
- 1 V = 1 J/c
- 1 kWh = 3.6 × 10⁶ J
When the applied electromotive force is 5.00 V, the energy required to produce 3.00 kg (3.00 × 10³ g) of aluminum is:

they move around and eventually come and join to make a solid