Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Abiotic factors, also called abiotic components, are non-living parts of the environment which affects the living organisms and their function in the ecosystem. I think the correct answer is A. Wind factor is the only non-living part from the choices.
The mass number plays an important role for elements and their isotopes. Mass number comes from the addition of protons and neutrons (their weight). Isotopes are the elements, but with a different number of neutrons. So in turn, by subtracting the number of protons (atomic number) from the mass, you can find the number of neutrons.
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:
Thus, we solve for the molar mass of the metal to obtain:
For the subsequent problems, we proceed as follows:
a.
b.
c.
Regards!
D. breakdown of rocks through mechanicals or chemicals processes
