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Answer:
Part A: 2N₂O(g) ⇄ 2N₂(g) + O₂(g)
Part B: -r = K*[N₂O]²
Part C: K= k1*k2
Explanation:
Part A
To do the balance chemical question for the overall chemical reaction, we must sum the reaction of the steps, eliminating the intermediaries, which are the compounds that have the same amount both at reactants and products (bolded).
N₂O(g) ⇄ N₂(g) + O(g)
N₂O(g) + O(g) ⇄ N₂(g) + O₂(g)
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2N₂O(g) + O(g) ⇄ 2N₂(g) + O(g) + O₂(g)
2N₂O(g) ⇄ 2N₂(g) + O₂(g)
Part B
The velocity of the reaction (r) can be calculated based on the reactants or based on the products. Let's do it based on the disappearing of the reactant. Because it is disappearing, the variation at its concentration must be negative, so the rate will be negative.
Let's suppose its an elementary reaction, so, the concentration of the reactant must be elevated by its coefficient. And let's call the overall rate constant as K:
-r = K*[N₂O]²
Part C
Because the steps were summed, and the reactions were not multiplied by a constant or inverted, the constant K is just the multiplication of the constants of the steps:
K= k1*k2
So it can break down the egg’s exterior and enter
Explanation:
The given data is as follows.
Mass flow rate of mixture = 1368 kg/hr
in feed = 40 mole%
This means that 
in feed = (100 - 40)% = 60%
We assume that there are 100 total moles/hr of gas 
in feed stream.
Hence, calculate the total mass flow rate as follows.
40 moles/hr of N_{2}/hr (28 g/mol of ) + 60 moles/hr of 
(2 g/mol of )
= 1120 g/hr + 120 g/hr
= 1240 g/hr
= 
(as 1 kg = 1000 g)
= 1.240 kg/hr
Now, we will calculate mol/hr in the actual feed stream as follows.
= 110322.58 moles/hr
It is given that amount of nitrogen present in the feed stream is 40%. Hence, calculate the flow of into the reactor as follows.
= 44129.03 mol/hr
As 1 mole of nitrogen has 28 g/mol of mass or 0.028 kg.
Therefore, calculate the rate flow of into the reactor as follows.
= 1235.612 kg/hr
Thus, we can conclude that the the feed rate of pure nitrogen to the mixer is 1235.612 kg/hr.
(a) 485 x 200 mg = 97000 mg of ibuprofen in the bottle
97000 mg x (1g/1000mg) = 97g of ibuprofen in the bottle
97g (1 mol/ 206.5gC13H18O2) = 0.46973 moles of ibuprofen in the bottle
(b) 0.46973 mol C13H18O2 (6.022 x 10^23 molecules/1mol) = 2.8287 x 10^23 molecules of ibuprofen in the bottle
