Part A The first step to engineering is to define the problem. Write down the problem the students have to solve, and describe t
2 answers:
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Answer:1) Volume of required is 55.98 mL.
2) 0.62577 grams of is produced.
Explanation:
1) Molarity of
Volume of
Molarity of
Volume of
According to reaction, 1 mole of reacts with 3 mole of
, then, 0.0041985 moles of
will react with:
moles of
that is 0.0125955 moles.
Volume of required is 55.98 mL.
2)
Number of moles of
According to reaction, 3 moles of gives 1 mole of
, then 0.004485 moles of
will give:
moles of
that is 0.001495 moles.
Mass of =
Moles of × Molar Mass of
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of is produced.
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°.
