when 100mL of 1.00 mol/L sodium hydroxide is added to 100mL of 1.00mol/L hydrochloric acid in an insulated container, the temperature rises from 21.0 degress to 34.6 degrees. Calculate the heat of neutralisation for the reaction.

Total q = mass H2O x specific heat H2O x delta T = 200g x 4.184 J/g x (34.6-21.0) = ?? (If you want this in calories substitute 1 cal/g for the 4.184 J/g.

This usually is expressed as J/mol, or cal/mol.

The heat of neutralization is the amount of heat released or absorbed when an acid and a base react and form a salt and water. To calculate the heat of neutralization, you need to calculate the heat released or absorbed during this reaction.

The temperature rise (ΔT) of the reaction is given as 34.6 - 21.0 = 13.6 degrees Celsius. We can convert this value to Kelvin by adding 273 to obtain 286.6K.

The heat released or absorbed (q) can be calculated using the equation q = m × c × ΔT, where q is the heat released/absorbed, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.

To calculate the heat of neutralization, let's assume that the specific heat capacity is equal to water (c = 4.18 J/g·K) and the density of the solution is equal to the density of water (ρ = 1 g/mL).

First, we need to calculate the total mass of the solution. In this case, since we are adding 100 mL of sodium hydroxide to 100 mL of hydrochloric acid, the total volume of the solution is 200 mL. Since the density of the solution is 1 g/mL, the total mass is 200 g. So, m = 200 g.

Next, we calculate the heat released or absorbed:

q = m × c × ΔT
= 200 g × 4.18 J/g·K × 13.6 K
≈ 113824 J

Therefore, the heat of neutralization for the reaction is approximately 113824 J.

Note: The heat of neutralization can vary slightly depending on experimental conditions, so the value calculated here is an approximation.