A100.0 ml sample of 0.300 m naoh is mixed with a 100.0 ml sample of 0.300 m hno3 in a coffee cup calorimeter. if both solutions were initially at 35.00°c and the temperature of the resulting solution was recorded as 37.00°c, determine the δh°rxn (in units of kj/mol nano3) for the neutralization reaction between aqueous naoh and hno3

-55.8 kJ/mol

Explanation:

There is a part missing from the question.

Assume no heat is lost to the calorimeter or the surroundings, and the density and heat capacity of the resulting solution are the same as water.

The initial moles of NaOH and HNO₃ are:

0.1000 L × 0.300 mol/L = 3.00 × 10⁻² mol

The neutralization reaction is:

NaOH + HNO₃ → NaNO₃ + H₂O

When 3.00 × 10⁻² moles of NaOH react with 3.00 × 10⁻² moles of HNO₃, they produce 3.00 × 10⁻² moles of NaNO₃ and 3.00 × 10⁻² moles of H₂O.

According to the law of conservation of energy, the sum of the heat released by the reaction and the heat absorbed by the solution is equal to zero.

ΔH°rxn + ΔH°sol = 0

ΔH°rxn = -ΔH°sol    [1]

The volume of the solution is 100.0 mL + 100.0 mL = 200.0 mL. Since the density is 1.00 g/mL, the mass of the solution is 200.0 g.

We can calculate the heat absorbed by the solution using the following expression.

ΔH°sol = c × m × ΔT = (4.184 × 10⁻³ kJ/g.°C) × 200.0 g × (37.00°C - 35.00°C) =  1.674 kJ

where,

c: specific heat capacity of the solution

m: mass of the solution

ΔT: change in the temperature

From [1],

ΔH°rxn = -1.674 kJ

We can express the enthalpy of reaction per mole of NaNO₃.

ΔH°rxn = -1.674 kJ / 3.00 × 10⁻² mol = -55.8 kJ/mol

precipitation

explanation:

when it goes from cloud to rain/back to earth in the form of rain