Answer:The molar heat of vaporization of molten lead is 178.03 kJ/mol.
Explanation:
The molar heats of fusion of lead
...(1)
The molar heats of sublimation of lead
...(2)
Adding (1) and (2)
The molar heat of vaporization of molten lead is 178.03 kJ/mol.
Answer:
168 g
Reaction:
Explanation:
First we need to figure out what is the formula mass (Mr) of Fe2O3 (Iron(III) oxide).
The atomic mass of Fe=56, O=16
Mr= 56+56+16+16+16=160 g/mol
We can now figure out the moles of the iron:
240/160 = 1.5 mol
Then we need to find the ratio to make Fe.
In the reaction, we know that 1 mole of Fe2O3 can make 2 moles of Fe. So 1.5 moles of Fe2O3 can make 3 moles of Fe.
And then we need to find the mass of the Fe.
3×56=168 g.
Hence 168g of the iron is made after the reaction.
Answer:
1 mol of water is produced in those conditions.
Explanation:
The reaction to produce water between H₂ and O₂ is this:
2H₂ + O₂ → 2H₂O
We don't have the amount of hydrogen, so we have to think that is in excess.
Let's work with oxygen.
Ratio is 1:2
For 0.5 mole of oxygen, I will make the double of moles of water.
Answer:
1. Molality.
2. Concentration.
3. Percent by mass.
4. Mole fraction.
Explanation:
Molarity is the measurement by which the number of moles is measured by a solute. The solution here is one liter respectively.
Concentration is the product which is emerged when the amount of the solute is divided by the amount o the solvent or solution.
Mole fraction refers to the number of component moles divided by the number of the solution moles.
Answer:
pH of the final solution = 9.15
Explanation:
Equation of the reaction: HCl + NH₃ ----> NH₄Cl
Number of moles of NH₃ = molarity * volume (L)
= 0.4 M * (300/1000) * 1 L = 0.12 moles
Number of moles of HCl = molarity * volume (L)
= 0.3 M * (175/1000) * 1 L = 0.0525 moles
Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced
Number moles of NH₄Cl produced = 0.0525 moles
Number of moles of base left unreacted = 0.12 - 0.0525 = 0.0675
pOH = pKb + log([salt]/[base])
pKb = -logKb
pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)
pOh = 4.744 + 0.109
pOH = 4.853
pH = 14 - pOH
pH = 14 - 4.853
pH = 9.15
Therefore, pH of the final solution = 9.15
