The acid dissociation constant(Ka) is 0.0095
The reaction for this dissociation of acid is
HSO4 ⇄H+ + SO4 -2
The dissociation constant can be determined from the following expression
[H+] = 10-PH
= 10-1.35
[H+] = 0.0447
[H+] = 0.0447 mol / L
From equation, [H+] = [SO4-2] = 0.0447 mol / L
[SO4-2] = 0.0447 mol / L
From the values of [H+], [SO4-2] and [HSO4] Ka can be calculated as follows,
Ka = 0.0447 * 0.0447 / 0.200
= 0.0019 / 0.200
= 0.0095
Hence the value of the acid dissociation constant (Ka) for the given reaction is 0.0095
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Answer:
B. Ca
Explanation:
Let's look at the electron configurations of all four elements (I am going to write it in noble gas configuration to make it simpler):
Mg electron configuration: [Ne]3s2
Ca electron configuration: [Ar]4s2
Ar electron configuration: [Ar]
K electron configuration: [Ar] 4s1
We notice that Ca has two electrons in the 4s sublevel, which satisfies what the question is asking for.
The answer is thus B. Ca.
We are given the number of moles:
O = 1.36 mol
H = 4.10 mol
C = 2.05 mol
To get the empirical formula, first divide everything by
the smallest number of moles = 1.36 mol. So that:
O = 1 mol
H = 3 mol
C = 1.5 mol
Next step is to multiply everything by a number such that
all will be a whole number. In this case, multiply by 2 to get a whole number
for C, so that:
O = 2 mol
H = 6 mol
C = 3 mol
Therefore the empirical formula of the compound is:
C3H6O2
francium , in the Periodic table the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. making helium is the smallest element, and francium the largest.