Please work out all the problems on this page, no links, no i don't knows please do not take advantage of me or my points please
actually answer these questions
2 answers:
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Answer:
1. 0.97 V
2.
Explanation:
In this case, we can start with the <u>half-reactions</u>:
With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:
I hope it helps!
Answer:
The number of atoms present in one unit of the following compounds is:
a). Potassium Iodide , KI = 2
b).Sodium Sulfide, = 3
c). Silicon Dioxide , = 3
d). Carbonic Acid , = 6
Explanation:
Atomicity : It is defined as the number of atoms that are present in a given molecule/compound.
Atom : The smallest unit of matter is called atom. For e.g O is atom of oxygen but is not an atom , it is molecule of oxygen .
molecule has 2 atoms of Oxygen
Similarly Na , K , Fe are atoms but ,
,
,
are molecules
a).Potassium Iodide
KI = 1 atom of K + 1 atom of I
Total atoms = 2
b) Sodium Sulfide
= 2 atoms of Na + 1 atom of S
Total atoms = 3
c) Silicon Dioxide
= 1 atom of Si + 2 atoms of O
Total atoms = 3
d) Carbonic Acid
= 2 atom of H + 1 atom of C + 3 atom of O
= 2+1+3
Total atoms = 6