The complete reaction is as,
4-Aminophenol + Acetic Anhydride → <span>Acetaminophen + Acetic Acid
First of all convert the ml of Acetic anhydrite to grams,
As,
Density = mass / volume
Solving for mass,
mass = Density </span>× Volume
<span>Putting values,
mass = 1.08 g/ml </span>× 5ml
<span>
mass = 5.4 g of acetic anhydride
First Find amount of acetic anhydride required to react completely with 2 g of p-Aminophenol,
As,
109.1 g of p-aminophenol required = 102.1 g of acetic anhydride
so, 2 g of p-aminophenol will require = X g of Acetic Anhydride
Solving for X,
X = (2 g </span>× 102.1 g) ÷ 109.1 g
X = 1.87 g of acetic anhydride is required to be reacted.
But, we are provided with 5.4 g of Acetic Anhydride, means p-aminophenol is the limiting reactant and it controls the formation of product. Now Let's calculate for product,
As,
109.1 g of p-aminophenol produced = 180.2 g of <span>Acetaminophen
So 2.00 g of p-aminophenol will produce = X g of Acetaminophen
Solving for X,
X = (2.00 g </span>× 180.2 g) ÷ 109.1 g
X = 3.30 g of Acetaminophen
Result:
<span>If 2.00g of p-aminophenol reacts with 5.00 ml of acetic anhydride 3.30 g of acetaminophen is made.</span>
Answer:
the answer would be 9,448.8
Answer:
B carbon
Explanation
Lewis structure or dot structure is an easy way to get the bonding details of atoms in a molecule. If we talk about methane molecule carbon is central atom with four electrons that are bonded to four hydrogen atoms and each bond is single covalent bond.
Please see attached figure,
Hope it helps!
Answer:
mL of NaOH required =29.9mL
Explanation:
Let us calculate the moles of vitamin C in the tablet:
The molar mass of Vitamin C is 176.14 g/mole
![moles=\frac{mass}{molarmass}=\frac{500mg}{176.14}=\frac{0.5}{176.14}=0.0028](https://tex.z-dn.net/?f=moles%3D%5Cfrac%7Bmass%7D%7Bmolarmass%7D%3D%5Cfrac%7B500mg%7D%7B176.14%7D%3D%5Cfrac%7B0.5%7D%7B176.14%7D%3D0.0028)
Thus we need same number of moles of NaOH to reach the equivalence point.
For NaOH solution:
![moles=MolarityXvolume=0.095Xvolume](https://tex.z-dn.net/?f=moles%3DMolarityXvolume%3D0.095Xvolume)
![0.00283=0.095Xvolume](https://tex.z-dn.net/?f=0.00283%3D0.095Xvolume)
![volume=0.0299L=29.9mL](https://tex.z-dn.net/?f=volume%3D0.0299L%3D29.9mL)