Answer:
FCI=88.0818 MM≅88 MM
Explanation:
Empirical correlation based on the work of Bridgwater and Mumford (1979):
For Liquid or solid phase Plants:
F<60,000 tonne/yr Eq (1)
F≥60,000 tonnes/yr Eq (2)
Where:
N is the number of functional units
F is the process throughput tonnes/yr
In our case F=40,000 tonne/yr <60,000 tonne/yr, We are going to use Eq (1)
F<60,000 tonne/yr
N=8, F=40,000 tonne/yr
FCI=88.0818 MM≅88 MM
Answer:
1) Increasing the pressure A) Shift to the left
2) Removing hydrogen gas B) Shift to the right
3) Adding a catalyst C) No effect
Explanation:
- <em>Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em></em>
<u><em>1) Decreasing the pressure:</em></u>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.
- So, decreasing the pressure will shift the reaction to the side with more moles of gas (left side).
<u><em>so, the right match is: A) Shift to the left.</em></u>
<em><u>2) Adding hydrogen gas:</u></em>
- Adding hydrogen gas will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of hydrogen gas by addition.
<u><em>so, the right match is: B) Shift to the right.</em></u>
<u><em></em></u>
<u><em>3) Adding a catalyst:</em></u>
- Catalyst increases the rate of the reaction without affecting the equilibrium position.
- Catalyst increases the rate via lowering the activation energy of the reaction.
- This can occur via passing the reaction in alternative pathway (changing the mechanism).
- The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
- in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
- with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
<u><em>so, the right match is: B) No effect.</em></u>
<u><em></em></u>
The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%
The quantum numbers are used to describe the position of an electron in atom.
<h3>What are quantum numbers?</h3>
The quantum numbers are used to describe the position of an electron in atom. I have attached a picture that shows the complete question.
For the 1s orbital;
n =1
l = 0
ml = 0
ms = ±1/2
For the 2s orbital;
n = 2
l = 1
ml = -1, 0, 1
ms = ±1/2
Learn more about quantum numbers:brainly.com/question/18835321
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Cutting the cardboard would be the first step
