Answer:
im sorry for being mean goodbye im get the real answer
Explanation:
True because hydrogen ions combines with h2o to make hydronium ion, so in a sense h2o is acting like a base.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure is ![[O_2] = 4.8 *10^{-5} \ atm](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%204.8%20%2A10%5E%7B-5%7D%20%5C%20atm)
Explanation:
From the question we are told that
The pressure of
is ![[SO_3 ] = 0.63 \ atm](https://tex.z-dn.net/?f=%5BSO_3%20%5D%20%3D%20%200.63%20%5C%20atm)
The pressure of
is ![[SO_ 2] = 0.30 \ atm](https://tex.z-dn.net/?f=%5BSO_%202%5D%20%20%3D%20%200.30%20%5C%20atm)
The equilibrium constant is
The reaction is
⇔ 
Generally the equilibrium constant is mathematically represented as

=> ![[O_2] = \frac{k_p * [SO_3] ^2 }{[SO_2]^2}](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%20%5Cfrac%7Bk_p%20%2A%20%5BSO_3%5D%20%5E2%20%7D%7B%5BSO_2%5D%5E2%7D)
substituting values
![[O_2] = \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%20%5Cfrac%7B1.2%20%2A10%5E%7B-5%7D%20%2A%200.60%20%5E2%20%7D%7B0.30%5E2%7D)
![[O_2] = 4.8 *10^{-5} \ atm](https://tex.z-dn.net/?f=%5BO_2%5D%20%3D%20%204.8%20%2A10%5E%7B-5%7D%20%5C%20atm)
The percentage of excess air used during combustion process of ethane will be 37 %.
Burning, also known as combustion, would be a high-temperature highly exothermic chemical process that occurs when an oxidant, typically atmospheric oxygen, interacts with a fuel to generate oxidized, frequently gaseous products in a mixture known as smoke.
Calculation of percentage of air .

Mair=Mair/Rin
+
÷
+
+
33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1
= 176/176+8
X= 0.37
0.37 × 100
X= 37%
Therefore, the percentage of excess air used during combustion process of ethane will be 37 %.
To know more about combustion process
brainly.com/question/13153771
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