Answer : The molar heat of solution of KCl is, 17.19 kJ/mol
Explanation :
First we have to calculate the heat of solution.
where,
q = heat produced = ?
c = specific heat capacity of water = 
= change in temperature = 0.360 K
Now put all the given values in the above formula, we get:
Now we have to calculate the molar heat solution of KCl.
where,
= enthalpy change = ?
q = heat released = 460.8 J
m = mass of 
= 2.00 g
Molar mass of = 74.55 g/mol
Now put all the given values in the above formula, we get:
Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol
Answer:
not sure if this is the right way to answer this question but PbS is Phosphate-buffered saline (i think)
Explanation:
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
