The answer to is all the information on a line graph is as precise as the information in the data table would be FALSE
Answer:
It is equal to Avogadro's number (NA), namely 6.022 x1023. If we have one mole of water, then we know that it will have a mass of 2 grams (for 2 moles of H atoms) + 16 grams (for one mole O atom) = 18 grams.
Explanation:
The question is not very much clear.
If you are asking for molecules then 1 mole water= 6.023 * 10^23
If you are asking for atoms then 1 mole water= 6.023 * 10^23 * 3
If you are asking for particles then,
So, in your example you would have one mole of water molecules. If you dissociated those water molecules, than you would end up with 2 moles of hydrogen atoms, and one mole of oxygen atoms.
I hope that was helpful!
H=1 proton,1 electron
O=8 protons,8 neutrons and 8 electrons
total particles in one H2O molecule-28
total no. of particles in 1 mole of water- 6.023 * 10^23 * 28
It looks all correct to me, great job!
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.