Answer: 16.32 g of 
as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of 
require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of 
Thus 16.32 g of as excess reagent are left.
Answer:
positive H and negative S
Explanation:
For a reaction to be spontaneous, the absolute best combination is a negative Delta H and a positive Delta S. When they are both positive, the reaction is only spontaneous at higher temperatures. When they are both negative, the reaction is only spontaneous at lower temperatures. and again if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed. The catalyst does not affect the energy of the reactants or products (and thus does not affect ΔG).
So from these discussions
Ea does not affect G value at all (whether +Ea or -Ea).
And for product to be formed the reaction should be spontaneous, where H is negative and S positive else the reaction will yield low product.
Answer: a) 
b) 
Explanation:
The reaction is :
Rate = Rate of disappearance of 
= Rate of appearance of 
Rate = ![-\frac{d[N_2O_5]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BN_2O_5%5D%7D%7B2dt%7D)
= ![\frac{d[NO_2]}{4dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7B4dt%7D)
Rate of disappearance of = 
= 
a) Rate of disappearance of =
Rate of appearance of =
b) Rate of appearance of = ![\frac{d[NO_2]}{dt}=2\times 1.7\times 10^{-4}}=3.4\times 10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D2%5Ctimes%201.7%5Ctimes%2010%5E%7B-4%7D%7D%3D3.4%5Ctimes%2010%5E%7B-4%7D)
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:
