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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The average atomic mass of the imaginary element : 47.255 amu
<h3>Further explanation </h3>
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% ..
isotope E-47 47.011 amu, 87.34%
isotope E-48 48.008 amu, 6.895
isotope E-49 50.009 amu, 5.77%
The average atomic mass :
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
Answer:
4
Explanation:
both are gases, both are compounds , they are nonmetal so that leaves 4
P1V1=P2V2
(1.00atm)(3.60L)=(2.50atm)V
3.60=(2.50atm)V
3.60/2.50=V
1.44L=V
