Answer:
<h2>12 atm</h2>Explanation:
To find the initial pressure we use the formula for Boyle's law which is
Since we are finding the initial pressure
From the question we have
We have the final answer as
<h3>12 atm</h3>Hope this helps you
How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal?
1 answer:
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Answer:
The answer to your question is the letter B
Explanation:
I will draw the skeletal structures of these compounds to determine which alcohol is secondary.
Secondary alcohol is alcohol in which the hydroxyl group is attached to a secondary alcohol.
Letter A has primary and secondary alcohol so I discard this choice.
Letter B has secondary alcohol, so this is the correct choice.
The letter C has a primary and 2 secondary alcohols so I discard this choice.
Answer:
7.3
Explanation:
By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that , and pKa = -logKa. Ka is the equilibrium constant of the acid.
Henderson Hasselbalch equation :
Where [HA] is the concentration of the acid, and is the concentration of the anion which forms the acid.
So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X
n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol
When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same stoichiometry.
Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of will be
n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol
So, the 0.0075 mol of reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of
, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.
The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.
Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!
So,
6.72 = pKa - log 4
pKa - log4 = 6.72
pKa = 6.72 + log4
pKa = 6.72 + 0.6
pKa = 7.3