Answer:
C₆H₆
Explanation:
We need to find the molecular formula of a compound of carbon (C) and hydrogen (H), so what <em>we need to find out is the number of atoms of C and of H in the molecule.</em> We know:
- molar mass = 78.1 g/mol
- C% = 92.3% = 92.3 g C / 100 g compound
So, in 1 mol of compound, 92.3% of the mass corresponds to Carbon:
<u>mass of C / mol of compound</u> = molar mass × C% = 78.1 g/mol × 92.3/100 = <u>72.1 g/mol</u>
<u>moles of C</u> = mass C / molar mass C = 72.1 g / 12.011 g/mol
moles of C = 6 moles of C per mol of compound
If 72.1 g in a mol of compound are Carbon atoms, the difference between the molar mass and the mass of Carbon atoms will correspond to H atoms in 1 mol of compound:
<u>mass of H / mol of compound</u> = molar mass - mass of C/mol
mass of H = 78.1 g / mol - 72.1 g /mol = <u>6.0 g/mol of compound</u>
<u>moles of H</u> = mass H / molar mass H = 6.0 g / 1.008 g/mol
moles of H = 6.0 moles of H per mol of compound
<em>So</em><em> one mol of compound has 6 moles of C and 6 moles of H.</em>
The molecular formula is then written as C₆H₆
By adding catalyst e.g. mno2,vanadium oxide
by increasing the temperature if it is an exothermic reaction
by also decreasing if it is an endothermic reaction
The ideal gas laws may be collectively expressed mathematically as:
PV/T = constant
Therefore, we may use this to determine the final pressure by equating the initial and final conditions:
PV/T = P * 2V / 2T
PV/T = PV/T
P = P
Therefore, the pressure is unchanged. It has stayed the same.
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:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)




Note:. The pH <7, then we have an acidic solution.