Answer:
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Answer:
Percent yield = 84.5 %
Explanation:
Given data:
Mass of methanol = 229 g
Actual yield of water = 219 g
Percent yield of water = ?
Solution:
Chemical equation:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
Number of moles of methanol:
Number of moles = mass/ molar mass
Number of moles = 229 g/ 32 g/mol
Number of moles = 7.2 mol
Now we will compare the moles of water with methanol.
CH₃OH : H₂O
2 : 4
7.2 : 4/2×7.2 = 14.4 mol
Mass of water:
Mass = number of moles × molae mass
Mass = 14.4 mol × 18 g/mol
Mass = 259.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 219 g / 259.2 g × 100
Percent yield = 84.5 %
Answer: pH = 3.15
Explanation: Solved in the attached picture.
<u>Answer:</u> The equilibrium concentration of HCl is 
<u>Explanation:</u>
We are given:
Moles of 
= 0.564 moles
Volume of vessel = 1.00 L
Molarity is calculated by using the equation:
Molarity of 
The given chemical equation follows:
<u>Initial:</u> 0.564
<u>At eqllm:</u> 0.564-x x x
The expression of 
for above equation follows:
![K_c=[NH_3][HCl]](https://tex.z-dn.net/?f=K_c%3D%5BNH_3%5D%5BHCl%5D)
The concentration of pure solid and pure liquid is taken as 1.
We are given:
Putting values in above equation, we get:
Negative sign is neglected because concentration cannot be negative.
So, ![[HCl]=2.26\times 10^{-3}M](https://tex.z-dn.net/?f=%5BHCl%5D%3D2.26%5Ctimes%2010%5E%7B-3%7DM)
Hence, the equilibrium concentration of HCl is
