Explanation:
You may not realise it, but you come across aldehydes and ketones many times a day. Take cakes and biscuits, for example. Their golden, caramelised crust is formed thanks to the Mailliard reaction. This is a process that occurs at temperatures above 140° C, when sugars with the carbonyl group in foods react with nucleophilic amino acids to create new and complex flavours and aromas.
Another example is formaldehyde. Correctly known as methanal, it is the most common aldehyde in industry. It has multiple uses, such as in tanning and embalming, or as a fungicide. However, we can also react it with different molecules to make a variety of more useful compounds. These include polymers, adhesives and precursors to explosives. But how do aldehydes and ketones react, and why?You should remember from Aldehydes and Ketones that they both contain the carbonyl functional group , . This is a carbon atom joined to an oxygen atom by a double bond. Let's take a closer look at it.
If we compare the electronegativities of carbon and oxygen, we can see that oxygen is a lot more electronegative than carbon.
Answer:Any change that occurs without altering the chemical composition of a substance is a physical change. Physical changes can include changing the color, shape, state of matter, or volume of a substance. It is crucial to remember that physical changes never alter the chemical makeup.
Explanation:
i hope that helps u try to figure it out a little bit sorry i couldn't find your answer i didn't have much to go off of
So,
Formate has a resonating double bond.
In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />
Answer:
Volume of stock solution needed = 6.0299 mL
Explanation:
<u>
</u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.
This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.
<u>Data:</u>
M1 = 6.01 M stock solution concentration
M2 = 0.3624 M diluted solution concentration
V2 =100 mL diluted solution volume
V1 = ? stock solution volume
M1 * V1 = M2 * V2
Answer:
The answers to the question are
1. 2nd and above order order
2. 2nd order
3. 1/2 order
4. 1st order
5. 0 order
Explanation:
We have 
1. For nth order reaction half life 
∝ ![\frac{1}{[A_{0} ]^{n-1} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%20%5D%5E%7Bn-1%7D%20%7D)
Therefore for a 0 order reaction increasing concentration of the reactant there will increase
First order reaction is independent [A₀].
Second order reaction [A₀] decrease, increase.
Similarly for a third order reaction
1. 2nd order
2. 2nd order reaction
3. Order of reaction is 1/2.
4. 1st order reaction.
5. Zero order reaction.
