Who was Dmitri Mendeleev? And why was he so important?
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Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:
From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:
= 126 L
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:
= 60.0 L
Similarly, let's calculate the volume of water vapors formed:
= 90.0 L
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:
= 30.0 L
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
Answer:
The solution 0.010 M has a higher percent ionization of the acid.
Explanation:
The percent ionization can be found using the following equation:
Since we know the acid concentration in the two cases, we need to find [H₃O⁺].
By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:
1. Case 1 (0.1 M):
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq) (1)
0.1 - x x x
(2)
Where:
Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.
By solving the above equation for x we have:
x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]
Hence, the percent ionization is:
2. Case 2 (0.01 M):
The dissociation constant from reaction (1) is:
With [CH₃COOH] = 0.01 M
By solving the above equation for x:
x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]
Then, the percent ionization for this case is:
As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.
Therefore, the solution 0.010 M has a higher percent ionization of the acid.
I hope it helps you!