How many joules are absorbed when 86.8 grams of water is heated from 40.0 °c to 86.0 °c?
11,400 joules
12,600 joules
14,100 joules
16,700 joules

16700Joules

Explanation:

Given parameters:

Mass of water = 86.8grams = 0.0868Kg

Initial temperature = 40°C, to kelvin, 40+273 = 313K

Final temperature = 86°C to kelvin, 86 + 273 = 359k

Solution:

This problem is to find the heat quantity Q that would be required to raise the temperature of the water through the given temperature.

     Q = mc(T₂ - T₁)

where m is the mass of the water

          c is the specific heat capacity of water = 4200JKg⁻¹K⁻¹

         T₂ and T₁ are the final and initial temperatures respectively

Q = 0.086 x 4200 x (359-313) = 16,615.2Joules or 16700Joules

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :