Answer:
Too high a value
Explanation:
HA + NaOH ⟶ NaA +H₂O
If the student has gone slightly past the equivalence point, they have added too much base.
The moles of HA are directly proportional to the moles of NaOH, so the moles of acid that the student calculates will be too high.
The calculated concentration of acid will also be too high.
Answer:
C. aluminum, fluorine, silicon
Explanation:
The absolute temperature is the lowest possible temperature in the universe. At this temperature, all atoms become motionless and cease to move.
The value of the absolute zero is pegged at -273.16°C.
- It is the lowest limit of the coldness of a body.
- Nothing can be colder than a body at absolute temperature.
- Many researches are underway to take advantage of this temperature value for scientific purpose.
- Thermodynamically, all process stops at this temperature.
- When a body is brought close a body at absolute zero, it can suffer cryogenic burn due to heat transfer.
B) proton
electrons are in a cloud around it, ions are elements with a charge, idk what a free radical is but ik that protons and neutrons are inside the nucleas
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.
