Given:
Mass of the ionic compound = 10.00 g
Mass of water = 75.0 g
Initial temperature of water T1= 23.2 C
Final temperature of water T2 = 31.8 C
Specific heat of water c = 4.18 J/gC
To determine:
Enthalpy of dissolution of the ionic compound
Heat gained by water equation:
Q = mcΔT
m = mass of water
c = specific heat
ΔT = change in temperature (T2-T1)
Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J
Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution)
Therefore, q(ionic) = 2696 J
ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g
Answer: A) enthalpy change = 2.7*10² J/g
Answer:
3.816 × 10⁻³ M
Explanation:
A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?
We can derive the following relations:
- The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
- Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.
The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:
The molarity of Cu²⁺ is:
The unknown acid is identified as either butanoic acid or ascorbic acid. To ascertain the number of moles based on the given molarity, we utilize the following relationship: Molarity of NaOH solution = 0.570 M and Volume of solution = 39.55 mL. Utilizing the values in the provided equation, we derive the necessary data. The equation governing NaOH and monoprotic acid reactions indicates that one mole of NaOH reacts with one mole of HX, resulting in 0.0225 moles of the monoprotic acid. Conversely, in the case of NaOH and diprotic acid interactions, the stoichiometry is such that two moles of NaOH engage with one mole of diprotic acid. Consequently, we can calculate moles for butanoic acid with a mass of 2.002 g and a molar mass of 88 g/mol, leading us to the conclusion that both butanoic and ascorbic acids represent the unknown acid being neutralized.
