Answer:
Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:
So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!
The mass of nucleons (and thus of the nucleus) is roughly 1000 times greater than that of electrons.
Warmer air is less dense than cold air.As air warm it rises while the cold air sink. Warmer air masses forces the cooler air to move which causes wind. These is illustrated when you open a hot oven the hotter air inside the oven rises into cooler air outside the oven.
<u>Analysing the Question:</u>
We are given a 250 mL solution of 0.5M K₂Cr₂O₇
Which means that we have:
0.5 Mole in 1L of the solution
0.125 moles in 250 mL of the solution <em>[dividing both the numbers by 4]</em>
<em />
<u>Mass of K₂Cr₂O₇ in the given solution:</u>
Molar mass of K₂Cr₂O₇(Potassium Dichromate) = 194 g/mol
<em>we know that we have 0.125 moles in the 250 mL solution provided</em>
Mass = Number of moles * Molar mass
Mass = 0.125 * 194
Mass = 36.75 grams
Answer:
Mass = 0.32 g
Explanation:
Given data:
Mass of CH₄ = ?
Volume of CH₄ = 500 mL (500 mL× 1L/1000 mL= 0.5 L)
Temperature = 273 K
Pressure = 1 atm
Solution:
Volume of CH₄:
500 mL (500 mL× 1L/1000 mL= 0.5 L)
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm× 0.5 L = n×0.0821 atm.L/ mol.K × 273 K
0.5 atm.L = n×22.4 atm.L/ mol
n = 0.5 atm.L / 22.4 atm.L/ mol
n = 0.02 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 0.02 mol × 16 g/mol
Mass = 0.32 g
