Answer:
10.00 mL
Explanation:
Please, do not forget to write the dot and 2 of Zeros after the number 10 in the answer!
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
Explanation:
a) For diatomic gas: Translational motion = 3 and rotational motion = 2
∴ Total (internal energy) = 3 + 2 = 5
b) Translational + Rotational + Vibrational = 3 + 2 + 1 = 6
c) Linear molecule
i) Non linear molecule
ii) Monatomic molecule
The answer is False. the amplitude shows how high or low something is
Electronic configuration of the atom describes the arrangemnet of electrons in different shells and subshells ( sublevels).
Now , there are 4 types of sublevels: s, p , d and f . These sublevels have orbital which are spaces with high probability of having an electron and each orbital can have maximum 2 electrons.
Therefore,
s-sublevel has 1 orbital - it can have maximum 2 electrons.
p-sublevel has 3 orbitals - it can have maximum 6 electrons
d-sublevel has 5 orbitals - it can have maximum 10 electrons
f-sublevel has 7 orbitals - it can have maximum 14 electrons.
Hence, the acsending order of sublevels in terms of maximum number of electrons is:
<h2>s < p < d < f</h2>
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.