Answer:
12.7g of Cu
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
Zn + CuSO4 —> ZnSO4 + Cu
Molar Mass of Cu = 63.5g/mol
Molar Mass of CuSO4 = 63.5 + 32 + (16x4) = 63.5 + 32 + 64 = 159.5g/mol
From the equation,
159.5g of CuSO4 produced 63.5g of Cu.
Therefore, 31.9g of CuSO4 will produce = (31.9 x 63.5) / 159.5 = 12.7g of Cu
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:
Explanation:
Lithium is in group 1 so there's 1 outermost shell electron.
Oxygen is in group 6 so there's 6 outermost shell electron.
To react lithium with oxygen, we need 2 lithium and 1 oxygen
Each lithium transfer 1 electron to the oxygen and oxygen gain 2 electrons.
formed.
If there's 4 lithium oxide, 4 molecules will be there.
If you mean how many ATOMS are there, there will be 3×4=12 atoms.
Answer:
pH=2.3
Explanation:
One H2SO4 molecule produce two H+ ions.
So conc. of H+ = 2(0.0025)
=0.005
Now, pH=-log (conc. of H+)
=-log(0.005)
=2.3
