Calculate the final temperature of solution when 10 g of oxalic acid, (COOH)2(s), is dissolved in 110.0 mL of water that is initially at 26 C. The enthalpies of formation of oxalic acid in the solid and aqueous phases are -826.8 and -818.8 kJ/mol, respectively. The heat capacity of water is 4.184 J/g*K. Ignore the heat capacity of oxalic acid.

This is done the same way as the NaCl problem posted earlier; however, in this case the DHsoln is a negative number which means you make q a + number.

To calculate the final temperature of the solution, we will use the equation for heat transfer:

q = m * C * ΔT

where q is the heat transferred, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

First, we need to find the amount of heat transferred (q) when the oxalic acid dissolves in water. We can calculate this using the enthalpies of formation of the solid and aqueous phases:

ΔH = Σ(ΔH of products) - Σ(ΔH of reactants)

Given that the enthalpy of formation of oxalic acid in the solid phase is -826.8 kJ/mol and in the aqueous phase is -818.8 kJ/mol, we can calculate the change in enthalpy (ΔH):

ΔH = -818.8 kJ/mol - (-826.8 kJ/mol) = 8 kJ/mol

Next, we need to determine the number of moles of oxalic acid that are dissolved. For this, we need to calculate the molar mass of oxalic acid:

C = 12.01 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

(C = 12.01 * 2) + (O = 16.00 * 2) + (H = 1.01 * 4) = 36 + 32 + 4.04 = 72.04 g/mol

Now, we can find the number of moles:

moles = mass / molar mass = 10 g / 72.04 g/mol = 0.139 moles

Since we know the specific heat capacity of water is 4.184 J/g*K, we can calculate the heat transferred when the temperature changes:

q = m * C * ΔT

The mass of water can be calculated using the density:

density = mass / volume
1 g/mL = mass / 110.0 mL
mass = 110.0 g

q = 110.0 g * 4.184 J/g*K * (Tf - 26 C)

Now we can substitute the values in the equation and solve for Tf:

q = 0.139 moles * 8 kJ/mol + 110.0 g * 4.184 J/g*K * (Tf - 26 C)

Simplifying the equation:

0.139 moles * 8000 J/mol + 461.04 J/gK * (Tf - 26 C) = 110.0 g * 4.184 J/g*K * Tf

1112.0 J + 461.04 J/gK * Tf - 12007.04 J = 459.04 J/gK * Tf

7616.96 J = 2.632 J/gK * Tf

Tf = 7616.96 J / (2.632 J/gK) = 2899.14 gK

Therefore, the final temperature of the solution is approximately 2899.14 C.