Answer:
the % yield is 82%
Explanation:
Given the data in the question,
we know that;
Molar mass of benzil is 210.23 g·mol−1
Molar mass of dibenzyl ketone is 210.27 g·mol−1
Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1
Now,
2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole
2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole
3.0 g of tetraphenylcyclopentadienone = 3 / 384.5 = 0.0078 mole
Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone
percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%
= 0.82 × 100%
= 82%
Therefore, the % yield is 82%
The carboxyl group in the amino acid is always acidic
Answer:
0.001 mole of NaF.
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 100 mL
Molarity = 0.01 M
Mole of NaF =?
Next, we shall convert 100 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100 mL × 1 L / 1000 ml
100 mL = 0.1 L
Thus, 100 mL is equivalent to 0.1 L.
Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:
Volume of solution = 0.1 L
Molarity = 0.01 M
Mole of NaF =?
Molarity =mole /Volume
0.01 = mole of NaF / 0.1
Cross multiply
Mole of NaF = 0.01 × 0.1
Mole of NaF = 0.001 mole.
Thus, 0.001 mole of NaF is present in 100 mL of the solution.
