The percentage of yield was 777.78%
<u>Explanation:</u>
We have the equation,
Be
[s] + 2
HCl
[aq] → BeCl
2(aq] +
H
2(g] ↑ Be
(s] +
2
HCl
[aq] → BeCl
2(aq] +
H
2(g]
↑
To find the percent yield we have the formula
Percentage of Yield= what you actually get/ what you should theoretically get x 100
=3.5 g/0.45 g 100
= 777.78 %
The percentage of yield was 777.78%
<span>Among the benefits; enlarging so that you can see it visually, making something complex and hard to understand to simplify it. One draw back is that not entirely accurate because it is basic. Additionally; models help us to visualize smaller structures that are too small to see properly. However; they do not take all variables into account and thus they may be inaccurate. </span>
Answer: Option (d) is the correct answer
Explanation:
The energy present within the bonds of the atoms of a compound or molecule is known as chemical energy.
As this energy is present at a position that is, within the bonds hence it is a potential energy. So, whenever there occurs a chemical reaction the stored chemical energy is released.
As potential energy is the energy possessed because of position of a substance. And, kinetic energy is the energy present due to the motion of an object.
Therefore, we can conclude that chemical energy is a form of potential energy.
Answer:
16.27 g of CaCO3 are produced upon reaction of 45 g CaCN2 and 45 g of H2O.
Explanation:
Ca(CN)2 + 3H2O → CaCO3 + 2 NH3
First of all, let's find out the limiting reactant.
Molar mass Ca(CN)2.
Molar mass H2O: 18 g/m
Moles of Ca(CN)2: mass / molar mass
45 g / 92.08 g/m = 0.488 moles
Moles of H2O: mass / molar mass
45g / 18g/m = 2.50 moles
This is my rule of three
1 mol of Ca(CN)2 needs 3 moles of H2O
2.5 moles of Ca(CN)2 needs (2.5 . 3) / 1 = 7.5 moles
I need 7.5 moles of water, but I only have 0.488. Obviously water is the limiting reactant; now we can work on it.
3 moles of water __ makes __ 1 mol of CaCO3
0.488 moles of water __ makes ___ (0.488 . 1) / 3 = 0.163 moles
Molar mass CaCO3 = 100.08 g/m
Molar mass . moles = mass
100.08 g/m . 0.163 moles = 16.27 g
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>
