Answer:
![[NO]=\frac{k_{-1}}{k_1} [N_2O_2]](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7Bk_%7B-1%7D%7D%7Bk_1%7D%20%5BN_2O_2%5D)
Explanation:
Hello!
In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:
![r=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=r%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:
![0=-k_1[NO]+k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=0%3D-k_1%5BNO%5D%2Bk_%7B-1%7D%5BN_2O_2%5D)
Which can be also written as:
![k_1[NO]=k_{-1}[N_2O_2]](https://tex.z-dn.net/?f=k_1%5BNO%5D%3Dk_%7B-1%7D%5BN_2O_2%5D)
Then, we solve for the concentration of NO to obtain:
Best regards!
I am not sure but rocky bodies ranging from large to small in size
Answer:
if you’re looking for a balanced equation it would be:
Methanol +ethanoic acid ==> methyl ethanoate + water ( in the presence of concentrated sulfuric acid )
CH3OH + CH3COOH==> CH3COOCH3 + H2O
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But, if you were looking for what it would be called it would be:
ethanoic acid with methanol will produce methyl ethanoate.
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And the reaction between Methanol and Ethanoic acid is known as esterification.
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I wasn’t sure of your question so I answered the best way that I could. I hope this helped!
The correct answer is the second option. <span>Baking soda fizzing in vinegar is an example of a chemical change. This example is a chemical reaction which undergoes a chemical change since new substances are being formed the fizzing of the system represent that gas is being produced by the reaction.</span>
Answer:
Percentage composition = 14.583%
Explanation:
In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.
Percentage composition by mass of Nitrogen
Nitrogen = 14g/mol
In one mole of the compound;
Mass of Nitrogen = 1 mol * 14g/mol = 14g
Mass of compound = 1 mol * 96.0 g/mol = 96
Percentage composition of Nitrogen = (Mass of Nitrogen / Mass of compound) * 100
percentage composition = 14/96 * 100
Percentage composition = 0.14583 * 100
Percentage composition = 14.583%
