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<span>63.4 g/mol
First, let's determine how many atoms per unit cell in face-centered cubic.
There is 8 corners, each of which has 1 atom, and each of those atoms is shared between 8 other unit cells. So 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each of which has 1 atom that's shared between 2 unit cells. So 6*1/2 = 3 atoms per unit cell. So each unit cell has the mass of 1+3 = 4 atoms.
Since there is 1000 liters per cubic meter, the mass per liter is 8920 kg/1000 = 8.920 kg/L. Now the mass per unit cell is 8920 g * 4.72x10^-26 = 4.21024x10^-22 g per unit cell. The mass per atom is 4.21024x10^-22 g / 4 = 1.05256x10^-22 g/atom, Finally, multiply by Avogadro's number, getting 1.05256x10^-22 g/atom * 6.0221409x10^23 atom/mol = 63.38664625704 g/mol.
Rounding to 3 significant digits gives 63.4 g/mol.</span>
C3H8+ 5 O2 --> 3 CO2 + 4 H2O
44 g. --------> 72 g
33.3 g. --------> x

Answer: The theoretical yield of H2O is 54.5
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.