This happens in massive star which convert hydrogen into helium?
1 answer:
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Answer:
2-Butene
Explanation:
The first step is the <u>ionization</u> of the acid to produce the hydronium ion. Then the OH will attack this ion to produce a <u>charged species</u> that can be stabilized when <u>H2O is produced</u>.
Then an <u>elimination</u> takes place to produce the more <u>substituted alkene</u> 2-butene and the <u>hydronium ion</u> is gain produced.
Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol
M (molarity) = 0.010 M (Mol/L)
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) =
Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:
![[ H_{3} O^+] = M* \alpha](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%20M%2A%20%5Calpha%20)
![[ H_{3} O^+] = 0.010* 2.23*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%200.010%2A%202.23%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 0.0223*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%200.0223%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%202.23%2A10%5E%7B-5%7D%20%5C%3Amol%2FL)
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
log10(2.23) ≈ 0.34
pH = ?
![[ H_{3} O^+] = 2.23*10^{-5}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%202.23%2A10%5E%7B-5%7D)
Formula:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)
Solving:
Note:. The pH <7, then we have an acidic solution.
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol
M (molarity) = 0.010 M (Mol/L)
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) =
Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
log10(2.23) ≈ 0.34
pH = ?
Formula:
Solving:
Note:. The pH <7, then we have an acidic solution.