In the following structure, carbons (I),(2),(3) and (4) are classified respectively as
1 answer:
<u>Answer:</u>
Carbon (i) : quaternary carbon
Carbon (ii) : secondary carbon
Carbon (iii) : tertiary carbon
Carbon (iv) : secondary carbon
<u>Explanation:</u>
Carbons can be classified into 4 categories:
(1) Primary carbon : These are the atoms where the carbon atom is attached to one other carbon atom.
(2) Secondary carbon : These are the atoms where the carbon atom is attached to two other carbon atoms.
(3) Tertiary carbon : These are the atoms where the carbon atom is attached to three other carbon atoms.
(4) Quaternary carbon : These are the atoms where the carbon atom is attached to four other carbon atoms.
In the given structure:
Carbon (i) is attached to 4 further carbon atoms and hence, it is a quaternary carbon.
Carbon (ii) is attached to 2 further carbon atoms and hence, it is a secondary carbon.
Carbon (iii) is attached to 3 further carbon atoms and hence, it is a tertiary carbon.
Carbon (iv) is attached to 2 further carbon atoms and hence, it is a secondary carbon.
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Answer:
Sr(OH)₂ will be the limiting reagent.
Explanation:
First of all, you should know the following balanced chemical equation:
2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂
The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:
- strontium hydroxide: 2 moles
- carbonic acid: 2 moles
Now, you know the following masses of the elements:
- Sr: 87.62 g/mole
- O: 16 g/mole
- H: 1 g/mole
So the molar mass of strontium hydroxide is:
Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole
You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?
moles of hydroxide= 0.103 moles
On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?
moles of carbonic acid= 0.525 moles
Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?
moles of carbonic acid= 0.103 moles
But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, <u><em>Sr(OH)₂ will be the limiting reagent.</em></u>