Which of the following is NOT a pure substance?
2 answers:
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Here is the correct question
You mix 125 mL of 0.170 M CsOH with 50.0 mL of 0.425 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 20.20 °C before mixing to 22.17 °C after the reaction. What is the enthalpy of reaction per mole of ? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J/g · K. Enthalpy of reaction = kJ/mol
Answer:
75.059 kJ/mol
Explanation:
The formula for calculating density is:
Making mass the subject of the formula; we have :
mass = density × volume
which can be rewritten as:
mass of the solution = density × volume of the solution
= 1.00 g/mL × (125+ 50 ) mL
= 175 g
Specific heat capacity = 4.2 J/g.K
∴ the energy absorbed is = mcΔT
= 175 × 4.2 × (22.17 - 20.00) ° C
= 1594.95 J
= 1.595 J
number of moles of CsOH =
= 0.2125 mole
Therefore; the enthalpy of the reaction =
=
= 75.059 kJ/mol
Answer:
A) pH = 8.11
B) pH = 7.96
Explanation:
The pH of buffer is calculated using Henderson Hassalbalch's equatio, which is
pKa = -logKa = -log(3.0 ✕ 10⁻⁸)
pKa = 7.5
[salt] = 0.667
[Acid] = 0.214
A) If the solution is halved it means the concentration of both salt and acid will be the same.
The volume will change
Number of moles of acid = molarity X volume = 0.214 X 25 = 5.35 mmol
Number of moles of salt = molarity X volume = 0.667 X 25 = 16.68 mmol
moles of NaOH added = molarity X volume = 0.100 X 10 = 1 mmol
These moles of NaOH will react with acid to give same amount of salt and thus the moles of acid will decrease
New moles of acid = 5.35 - 1 = 4.35
New moles of salt = 16.68 + 1 = 17.68
the new pH will be
B)
moles of HCl added = molarity X volume = 0.100 X 10 = 1 mmol
These moles of HCl will react with salt to give same amount of acid and thus the moles of salt will decrease
New moles of acid = 5.35 + 1 = 6.35
New moles of salt = 16.68 - 1 = 15.68
the new pH will be