Answer:
648.68 mg
Explanation:
The reaction that takes place is:
First we<u> calculate how many moles of each reactant were added</u>, using the <em>given volumes and concentrations</em>:
24 mmol of FeCl₃ would react completely with (24 * 3) 72 mmol of NaOH. There are not as many NaOH mmoles, so NaOH is the limiting reactant.
Now we <u>calculate how many moles of Fe(OH)₃ are formed</u>, using the <em>moles of the limiting reactant</em>:
Finally we <u>convert 6.07 mmol Fe(OH)₃ to grams</u>, using its<em> molar mass</em>:
Answer and Explanation
The isomer picked is the N-Propylamine.
It has a lone pair of electron available on the electron rich Nitrogen and no formal charge.
Since it will be hard to draw the Lewis structure in this answer format, I'll attach a picture of the Lewis structure to this answer.
The lone pair of electron is shown by the two dots on the Nitrogen atom.
The question is incomplete, complete question is :
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?
Structure is given in an image?
Answer:
There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Explanation:
Total numbers of carbon = 10
Number of primary carbons that is carbon joined to just single carbon atom = 6
Number of secondary carbons that is carbon joined to two carbon atoms = 1
Number of tertiary carbons that is carbon joined to three carbon atoms = 2
Number of quartenary carbons that is carbon joined to four carbon atoms = 1
So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
The reducing agent can approach the carbonyl face of camphor by forming a one carbon bridge (known as an exo attack) or a two carbon bridge (termed endo).
The two resultant stereoisomers are known as isoborneol and borneol (from exo attack) (from endo attack). Gas chromatography (GC) analysis may be used to calculate the ratio of each isomeric alcohol in the mixture. Unfortunately, IR analysis does not permit this.
The stereochemistry of the reaction is regulated in stiff cyclic compounds like camphor and norcamphor by protecting one side of the carbonyl group from the reagent's assault. The hydrogen atom is added to the endo side, creating the exo alcohol isoborneol, while the methyl groups on the one-carbon bridge of camphor screen the approach of the hydride from the "top" or exo side of the two-carbon bridge. You will be asked to guess the main isomeric alcohol created by the norcamphor hydride reduction later in the lab report.
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