Answer:
Water
Explanation:
one molecule of water has two hydrogen atoms covalently bonded to a single oxygen atom. Water is a tasteless, odorless liquid at ambient temperature and pressure. Which is known to be crystal blue.
Answer:
4–ethyl–2,3–dimethylheptane
Explanation:
To name the compound given above, do the following:
1. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest continuous carbon chain is 7. Thus the parent name is Heptane.
2. Identify the substituent group attached to the compound. In this case, the substituent group attached are:
a. Methyl (–CH₃). There are two methyl group attached.
b. Ethyl (–CH₂CH₃)
3. Locate the position of the substituent group attached to the compound by naming alphabetically.
a. The two Methyl (–CH₃) groups are located at carbon 2 and 3
b. The Ethyl (–CH₂CH₃) is located at carbon 4.
NOTE: The position of the Ethyl (–CH₂CH₃) group is the same from both side so we consider the lowest count for the methyl group.
4. Combine the above to obtain the name of the compound.
The name of the compound is:
4–ethyl–2,3–dimethylheptane
Lecithin is an emulsifier agent that's composed of 5 smaller molecules: phosphoric acid, choline, glycerol( is the backbone), and two fatty acids.
The fatty acids, which are hydrophobic (afraid of water), makes this substance more similiar to fats and represent the non-polar part of the lecithin.
The phosphate group is the polar portion of the molecule and it's the negatively charged. The choline is positively charged, which readily dissolve in water<span>.
</span><span>Lecithin is a good emulsifier because of these structural features. the hydrophobic contacts with the oil, while the hydrophilic end contacts with the water.</span>
First step is to balance the amount of atoms on each side
So for example if my equation that I have to balance is : Br2 -> Br-
First thing I would have to do would be :
Br2 -> 2Br-
Then I would have to balance the charges on both sides. So because there is a -2 charge on the right side I will have to add a -2 charge onto the right side as well. I would do this by :
Br2 + 2e- -> 2Br-
And now you can see the amount of atoms and the charges are balanced... So your overall reaction is now balanced too.
