Calculate the heat of neutralization (kJ/mole) for the reaction of 50.0 mL of 0.177 M potassium hydroxide with 100.0 mL 0.841 M sulfuric acid resulting in a temperature increase of 4.61 °C. Assume the solutions have a density of water and specific heat capacity of water, 4.184 J/(g-C). The calorimeter is a perfect system with no heat loss. 2KOH + H,SO, K;SO, + 2H,0 -327 k/mole O +34,4 k/mole O +2890 kJ/mole O -34.4 kJ/mole O +109 k/mole O -218 kJ/mole O -2890 kJ/mole +218 kJ/mole -109 kl/mole -654 k/mole O +327 kJ/mole O +654 kJ/mole

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Calculate the heat of neutralization (kJ/mole) for the reaction of 50.0 mL of 0.177 M potassium hydroxide with 100.0 mL 0.841 M sulfuric acid resulting in a temperature increase of 4.61 oC.  Assume the solutions have a density of water and specific heat capacity of water, 4.184 J/(g⋅oC).  The calorimeter is a perfect system with no heat loss.

 

Calculate the heat of neutralization (kJ/mole) for the reaction of 50.0 mL of 0.177 M potassium
hydroxide with 100.0 mL 0.841 M sulfuric acid resulting in a temperature increase of 4.61 °C.
Assume the solutions have a density of water and specific heat capacity of water, 4.184 J/(g-°C).
The calorimeter is a perfect system with no heat loss.
2KOH + H2SO4→ K2SO4 + 2H20
O -327 kJ/mole
O +34.4 kJ/mole
O +2890 kJ/mole
O -34.4 kJ/mole
O +109 kJ/mole
O -218 kJ/mole
O -2890 kJ/mole
O +218 kJ/mole
O -109 kJ/mole
O -654 kJ/mole
O +327 kJ/mole
O +654 kJ/mole
Transcribed Image Text:Calculate the heat of neutralization (kJ/mole) for the reaction of 50.0 mL of 0.177 M potassium hydroxide with 100.0 mL 0.841 M sulfuric acid resulting in a temperature increase of 4.61 °C. Assume the solutions have a density of water and specific heat capacity of water, 4.184 J/(g-°C). The calorimeter is a perfect system with no heat loss. 2KOH + H2SO4→ K2SO4 + 2H20 O -327 kJ/mole O +34.4 kJ/mole O +2890 kJ/mole O -34.4 kJ/mole O +109 kJ/mole O -218 kJ/mole O -2890 kJ/mole O +218 kJ/mole O -109 kJ/mole O -654 kJ/mole O +327 kJ/mole O +654 kJ/mole
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