Answer:
The answer to your question is:
1.- CO
2.- 0.414 moles of CO2
Explanation:
Data
2CO + O2 ⇒ 2CO2
CO = 0.414 moles
O2 = 0.418
Process
theoretical ratio CO/O2 = 2/1 = 1
experimental ratio CO/O2 = 0.414/0.418 = 0.99
Then the limiting reactant is CO
2.-
2 moles of CO --------------- 2 moles of CO2
0.414 moles of CO --------- x
x = (0.414 x 2) / 2
x = 0.414 moles of CO2
1) The more mass is the more entropy , because there are more particles, there is disorder.
2) Than higher temperature --- the more entropy.
3) Gas has more disorder than liquid, so gas has more entropy.
So, correct answer is E.
<span>Equation:2H2(g) + O2(g) → 2H2O(g)
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Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.
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Yes. bromine and sodium iodide can react to form sodium bromine and free iodine
The molarity of the stock solution is 1.25 M.
<u>Explanation:</u>
We have to find the molarity of the stock solution using the law of volumetric analysis as,
V1M1 = V2M2
V1 = 150 ml
M1 = 0.5 M
V2 = 60 ml
M2 = ?
The above equation can be rearranged to get M2 as,
M2 = 
Plugin the values as,
M2 = 
= 1.25 M
So the molarity of the stock solution is 1.25 M.
