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)
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Answer:
Mass = 4.6 g
Explanation:
Given data:
Number of molecules of sucrose = 8.1 ×10²¹ molecules
Mass of sucrose = ?
Solution:
First of all we will calculate the number of moles by using Avogadro number.
1 mole × 8.1 ×10²¹ molecules / 6.022×10²³ molecules
1.35 × 10⁻² mol
Mass of sucrose:
Mass = number of moles × molar mass
Molar mass = 342.3 g/mol
Mass = 1.35 × 10⁻² mol ×342.3 g/mol
Mass = 462.1 × 10⁻² g
Mass = 4.6 g
Answer:
There's a total of 8 oxygen atoms in the formula Cu3(PO4)2
Cl2=3.17g/L
Ne=.901g/L
CO2=1.96g/l
therefore Cl2 is the densest gas under the given conditions.
