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:
The enthalpy of the second intermediate equation is altered by halving its value and changing the sign.
Explanation:
Let's examine both the first and second intermediate reactions alongside the overall equation concerning the examined process;
First reaction;
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second reaction;
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Thus, the overall reaction becomes;
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH =?
According to Hess's law, which states that the total heat change in a reaction is equal to the sum of the heat changes for each step, we cannot simply sum the enthalpies for this overall reaction. Instead, we obtain the overall enthalpy by halving the second intermediate reaction's enthalpy and changing its sign before adding, as illustrated below;
Enthalpy of Intermediate reaction 1 + ½(-Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction
To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature.
P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k
Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
