Answer:
The pKa of the conjugate acid is 17.7
Explanation:
If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,
Kₐ = [OH⁻]/[H₂O]
Now, we determine the equivalent pKa
pKa = -log[ka]
pKa = -log[100]
pKa = -2
Removal of hydrogen from water is reversible as shown below;
H₂O ⇄ OH⁻ + H⁺
15.7 -2
This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];
pKa of the conjugate acid = 15.7 - (-2) = 17.7
The pKa of the conjugate acid is 17.7
Number
The atomic number is the number of protons in one atom of an element.
Hope this helps!
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hey there!:
density = 75.0 g/mL
Volume = 12 mL
mass = ?
Therefore:
D = m / V
75.0 =m / 12
m = 75.0 * 12
m = 900 g
Answer B
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Answer:
The molarity of this final solution is 0.167 M
Explanation:
Step 1: Data given
Volume of a 0.100 M HNO3 solution = 50.0 mL
Volume of a 0.200 M HNO3 = 100.0 mL
Step 2: Calculate moles
The final molarity must lie between 0.1M and 0.2M
Moles = molarity * volume
Moles HNO3 in 50mL of a 0.100M solution = 0.05 L *0.100 M = 0.005 mol
Moles HNO3 in 100mL of a 0.200M solution = 0.100 L*0.200 = 0.020mol
total moles = 0.005+0.020 = 0.025 moles in 150mL solution = 0.150L
Step 3: Calculate molarity of final solution
Molarity = mol / volume
Molarity 0.025 moles /0.150 L
Molarity = 0.167M
The molarity of this final solution is 0.167 M