Answer:
Near the boiling point of the solvent
Explanation:
The process of recrystallization is hinged on the fact that the amount of solute that can be dissolved by a solvent increases with temperature. The process involves creation of a solution by dissolving a solute in a solvent at or near its boiling point. At the boiling point of the solvent, the solute has a greater solubility in the solvent; not much volume of the hot solvent is required to dissolve the solute.
Before the solution is later cooled, you can now filter out insoluble impurities from the hot solvent. The quantity of the original solute drops appreciably because impurities have been removed. At this lower temperature, the solution becomes saturated and the solute can no longer be held in solution hence it forms pure crystals of solute, which can be recovered.
Recrystallization must be carried out using the proper solvent. The solute must be relatively insoluble in the solvent at room temperature but more soluble in the solvent at elevated temperature.
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.
Answer:
1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Explanation:
1) Possible reactions
2Al + 6HCl ⟶ 2AlCl₃ + 3H₂
Fe + 2HCl ⟶ FeCl₂ + H₂
2) Mass of each metal
a) Mass of Cu
The waste was the unreacted copper.
Mass of Cu = 2.5 g
b) Masses of Al and Fe
We have two relations
:
Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g
H₂ from Al + H₂ from Fe = 6.38 L at NTP
i) Calculate the moles of H₂
NTP is 20 °C and 1 atm.
(ii) Solve the relationship
Let x = mass of Al. Then
7.5 - x = mass of Fe
Moles of Al = x/27
Moles of Fe = (7.5 - x)/56
Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x
/18
Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56
∴ x/18 + (7.5 - x)/56 = 0.2652
56x + 18(7.5 - x) = 267.3
56x + 135 - 18x = 267.3
38x = 132.3
x = 3.5 g
Mass of Al = 3.5 g
Mass of Fe = 7.5 g - 3.5 g = 4.0 g
The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g
Answer:
13.7 moles of O₂ are needed
Explanation:
In order to find the moles of reactants that may react to make the products we need to determine the reaction:
Reactants are hydrogen and oxygen
Product: Water
2 moles of hydrogen can react to 1 mol of oxygen and produce 2 moles of water.
Balanced reaction: 2H₂(g) + O₂(g) → 2H₂O(l)
If 2 moles of hydrogen need 1 mol of oxygen to react
Therefore, 27.4 moles of H₂ must need (27.4 .1) / 2 = 13.7 moles of O₂
Water polluting
increased invasive species
overfishing
electricity cuts
loss of homes
