<em>it is given as as under: :
</em>
Tiny organisms known as cyanobacteria, or blue-green algae.
Answer:
Explanation:
<u>1) Equilibrium equation (given):</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
<u>2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - x x x
<u>3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - 0.3485 0.348 0.348
<u>4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):</u>
- Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²
- A² = 56.0 / 0.348² = 462.
Answer:
See explanation.
Explanation:
Hello,
In this case, we say that chemical reactions are governed by the law of conservation of mass, which states that matter cannot be neither created nor destroyed by transformed, for that reason, we need to balance chemical reactions in order to ensure all the atoms to be in the same quantity at both reactants and products.
Moreover, equilibrium is defined as such condition at which the concentration of both reactants and products stop changing over the time so they become constant as well as their null reaction rate.
A widely acknowledged reaction is the HABER one which consists on the synthesis of ammonia by using elemental nitrogen and hydrogen:

In such reaction, we have two nitrogens at both reatants and products and six hydrogens at at both reatants and products for us to obey the law of conservation of mass. Furthermore, as the time goes by, nitrogen reacts with hydrogen, nonetheless, they do not react indefinitely, they have a limit that is equilibrium, so their moles stop being consumed and remain unchanged as well as the produced moles of ammonia.
Best regards.
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!