Answer:
A
Explanation:
Hmm, so we have the following in the diagram
Pt(s)
Cl2(g)
Ag(s)
NaCl(aq)
AgNO3(aq)
Pt 2+, 4+, 6+ Though it states Pt is inert
Cl 2-
Ag 1+
Na 1+
NO3-
Anode definition: the positively charged electrode by which the electrons leave an electrical device.
Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.
Cations attracted to cathode pick up electrons
Anions attracted to anode release electrodes+
Reduction at Cathode (red cat gain of e)
Oxidation at Anode (ox anode loss of e)
So from the diagram we can see that the charge is being generated through the 2 metal plates.
So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-
Answer:- The Ka for the acid is
.
Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
Now, we make the ice table for this equation as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
I 0.25 0 0
C -X +X +X
E (0.25 - X) X X
where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.
X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.
![Ka = [H^+][A^-]\frac{1}{HA}](https://tex.z-dn.net/?f=Ka%20%3D%20%5BH%5E%2B%5D%5BA%5E-%5D%5Cfrac%7B1%7D%7BHA%7D)
Where, Ka is the acid ionization constant. Let's plug in the values.

Let's calculate the value of X first using the equation:
[/tex]
on taking antilog ob above equation we get:
![[H^+]=10^-^p^H](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E-%5Ep%5EH)
![[H^+]=10^-^2^.^7^1](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E-%5E2%5E.%5E7%5E1)
= 0.00195
So, X = 0.001195
Let's plug in this value of X in the equation:-


So, the value of Ka for butyric acid is
.
Answer:
The new pressure will be 0.225 kPa.
Explanation:
Applying combined gas law:

where,
are initial pressure and volume at initial temperature
.
are final pressure and volume at initial temperature
.
We are given:



Putting values in above equation, we get:


Hence, the new pressure will be 0.225 kPa.
Answer:
1: [H+] = 0.01 M
2: [H+] = 0.0001 M
3: [H+] = 0.0001 M
Explanation:
Step 1: data given
pH = -log[H+]
pH = pOH = 14
Step 2:
1. A solution with pH = 2.0
pH = 2
-log[H+] = 2.0
[H+] = 10^-2
[H+] = 0.01 M
2. A solution with pH = 4.0
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M
3. A solution with pOH = 10.0
pH = = 14 - 10 = 4
pH = 4
-log[H+] = 4.0
[H+] = 10^-4
[H+] = 0.0001 M