After ionization, sodium gains a net positive charge cuz sodium loses its 1 valence electron to gain the nearest stable octet which is neon{Ne}. Hope it helps
Answer: Option (d) is the correct answer.
Explanation:
Steps involved for the given reaction will be as follows.
Step 1:
(fast)
Rate expression for step 1 is as follows.
Rate = k ![[NO]^{2}](https://tex.z-dn.net/?f=%5BNO%5D%5E%7B2%7D)
Step 2: 
This step 2 is a slow step. Hence, it is a rate determining step.
Step 3.
(fast)
Here,
is intermediate in nature.
All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.
Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.
When Sodium and Chlorine come together they transfer an electron.
- Source: google
Hopefully this was clear and you understood!
2H(+) + SO4(2-) + Ca(2+) + 2I(-) -> CaSO4(s) + 2H(+) + 2I(-)
The signs in brackets are the subscripts for the charge of the ion. This is the complete ionic equation. The net ionic equation is:
Ca(2+) + SO4(2-) -> CaSO4
<h2>

→

</h2>
Explanation:
Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.
→
[oxidation by loss of hydrogen]
-
An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.
- An oxidizing agent used along with dilute sulphuric acid for acidification.
Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.
- The oxidation of aldehydes to carboxylic acids can be done by the two-step process.
- In the first step, one molecule of water is added in the presence of a catalyst that is acidic.
- There is a generation of a hydrate. (geminal 1,1-diol).
→
[reduction by the gain of electrons]
Here, the oxidizing agent used is
in the presence of acetone.