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:
![3.00 \times 10^{3} gAl.\frac{1molAl}{26.98gAl} .\frac{6mole^{-}}{2molAl}.\frac{96468c}{1mole^{-}}.\frac{5.00J}{c}.\frac{1kWh}{3.6 \times 10^{6}J} =44.7kWh](https://tex.z-dn.net/?f=3.00%20%5Ctimes%2010%5E%7B3%7D%20gAl.%5Cfrac%7B1molAl%7D%7B26.98gAl%7D%20.%5Cfrac%7B6mole%5E%7B-%7D%7D%7B2molAl%7D.%5Cfrac%7B96468c%7D%7B1mole%5E%7B-%7D%7D.%5Cfrac%7B5.00J%7D%7Bc%7D.%5Cfrac%7B1kWh%7D%7B3.6%20%5Ctimes%2010%5E%7B6%7DJ%7D%20%3D44.7kWh)
Answer:
5 cm³.
Explanation:
The following data were obtained from the question:
Volume of water = 35 mL
Volume of water + Rock = 40 mL
Volume of rock =.?
The volume of the rock can be obtained as follow:
Volume of rock = (Volume of water + Rock) – (Volume of water)
Volume of rock = 40 – 35
Volume of rock = 5 mL
Finally, we shall convert 5 mL to cm³. This can be obtained as illustrated below:
1 mL = 1 cm³
Therefore,
5 mL = 5 cm³
Thus, the volume of the rock is 5 cm³.