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.
The direction of heat flow is increased which means blocks temperature is higher and hotter than it was before
Endocytosis is a type of bulk transport which involves the movement of large particles through the membrane in and out
Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms
<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>
<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>
<span>total moles = .57 + 3.72 = 4.29 moles </span>
<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>
<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>
<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>
<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>
<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
140 g of nitrogen (N₂)
Explanation:
We have the following chemical equation:
N₂ + 3 H₂ -- > 2 NH₃
Now, to find the number of moles of ammonia we use the Avogadro's number:
if 1 mole of ammonia contains 6.022 × 10²³ molecules
then X moles of ammonia contains 6.022 × 10²⁴ molecules
X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³
X = 10 moles of ammonia
Taking in account the chemical reaction we devise the following reasoning:
If 1 mole of nitrogen produces 2 moles of ammonia
then Y moles of nitrogen produces 10 moles of ammonia
Y = (1 × 10) / 2
Y = 5 moles of nitrogen
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of nitrogen (N₂) = 5 × 28 = 140 g
Learn more about:
Avogadro's number
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