Answer:
The reaction will continue in the forward direction until all the NO or all the NO₂ is used up.
Explanation:
- <em>Le Châtelier's principle </em><em>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>
- So, according to Le Chatelier's principle, removing the product (N₂O₃) from the system means decreasing the concentration of the products; thus, the reaction will proceed forward to produce more product to minimize the stress of removing N₂O₃ from the system.
- <em>So, the reaction will continue in the forward direction until all the NO or all the NO₂ is used up.
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Methane is the compound CH4, and burning it uses the reaction:
CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.
Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.
n = 9.5/16.042 = 0.592195 mol
Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>
Friction is the force that slows down a ball, if their were no friction, it would not stop.
Answer: friction
