A 100.2 mL sample of 1.00 M NaOH is mixed with 50.1 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup; the cup is fitted with a lid through which passes a calibrated thermometer. The temperature of each solution before mixing is 22.45 °C. After adding the NaOH solution to the coffee cup and stirring the mixed solutions with the thermometer, the maximum temperature measured is 30.90 °C. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g·°C), and that no heat is lost to the surroundings. Write a balanced chemical equation for the reaction that takes place in the Styrofoam cup. Remember to include phases in the balanced chemical equation. Calculate the enthalpy change per mole of H2SO4 in the reaction.

∆H = 95.6 kJ/mol

Explanation:

Step 1: Data given

Volume of a 1.00 M NaOH = 100.2 mL

Volume a 1.00 M H2SO4 = 50.1 mL

The temperature of each solution before mixing is 22.45 °C

The maximum temperature measured is 30.90°C

the density of the mixed solutions is 1.00 g/mL

The specific heat of the mixed solutions is 4.18 J/g°C

Step 2: The balanced equation

2NaOH(aq) + H2SO4(aq) ==> Na2SO4(aq)+ 2H2O(l)

Step 3:

q = mC∆T

⇒q = the heat transfer = TO BE DETERMINED

⇒m = the mass of solution = 100.2 mL + 50.1 mL = 151.2 mL

           151.3 ml * 1g/mL = 151.3 grams

⇒c is the specific heat of the solution = 4.18 J/g°C

⇒∆T = the change in temperature = T2 - T1 = 30.90 °C - 22.45 °C = 8.45 °C

q = 151.3 grams * 4.18 Jg°C * 7.65 °C

q = 4838.1 J

Step 4: Calculate ∆H per mole H2SO4

moles H2O4 = 0.0501 L*1.00 M = 0.0501 moles

∆H =4838.1 J / 0.0501 moles = 95569 J/mole = 95.6 kJ/mol

the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).

the answer is option a.