In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.00mm away. What is
1 answer:
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Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of =
= 0.0458 M
Concentration of =
= 0.0521 M
GIven that :
Ka =
Thus; it's pKa = 4.72
pH ≅ 4.80
Answer:
\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}
Explanation:
Hello,
The suitable differential equation for this case is:
As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:
Of course, .
Now, since the volume of a cone is and the ratio
or
, the volume becomes:
We proceed to its differentiation:
Then, we compute
Finally, at h=2:
Best regards.