To know the acidity of a
solution, we calculate the pH value. The formula for pH is given as:
<span>pH = - log [H+] where H+ must be in Molar</span>
We are given that H+ = 3.25 × 10-2 M
Therefore the pH is:
pH = - log [3.25 × 10-2]
pH = 1.488
Since pH is way below 7, therefore the solution
is acidic.
To find for the OH- concentration, we must
remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,
[H+]*[OH-] = 10^-14 <span>
</span>[OH-] = 10^-14 / [H+]
[OH-] = 10^-14 / 3.25 × 10-2
[OH-] = 3.08 × 10-13 M
Answers:
Acidic
[OH-] = 3.08 <span>× 10-13 M</span>
Answer:
4.285 L of water must be added.
Explanation:
Hello there!
In this case, for this dilution-like problems, we need to figure out the final volume of the resulting solution so that we would be able to obtain the correct volume of diluent (water) to be added. In such a way, we can obtain the final volume, V2, as shown below:

Thus, by plugging in the initial molarity, initial volume and final molarity (0.587 M) we obtain:

It means we need to add:

Of diluent water.
Regards!
Answer:
Answer is given below.
Explanation:
Anode is that electrode where oxidation occurs. Cathode is that electrode where reduction occurs.
In cell representation, half cell present left to salt-bridge notation
is anodic system and another half cell present right to salt-bridge notation
is cathodic system.
So anode is Cu and cathode is Ag.
oxidation: 
[reduction:
]
-----------------------------------------------------------------------------------------------
chemical equation: 
Oxidizing agent is that species which takes electron from another species. Here
takes electron from Cu. Hence
is the oxidizing agent.
Reducing agent is that species which gives electron to another species. Here Cu gives electron to
. Hence Cu is the reducing agent.
<h3>Ferric Oxide:-</h3>


<h3><u>C</u><u>a</u><u>l</u><u>c</u><u>i</u><u>u</u><u>m</u><u> </u><u>H</u><u>y</u><u>d</u><u>r</u><u>o</u><u>x</u><u>i</u><u>d</u><u>e</u><u>:</u><u>-</u></h3>

