If you mix 100 milliliters of 10 degrees celsius of water and 100 milliliters of 90 degrees celsius of water what will be your final temperature of the mixture

Final temperature of the mixture is 50°C

Explanation:

This is a typical calorimetry problem,  where the heat yielded by a compound is immediately won by another.

Water at  90°C raise the temperature of water at 10°

Density of water = 1g/mL

Mass of water = 100 g, because the volume is 100mL

If the two pieces of water are in a calorimeter where heat is not released or gained, the sum of both heats is 0

4.18 J/g°C . 100 g ( T°final - 90°C) +  4.18 J/g°C . 100 g ( T°final - 10°C) = 0

418 J/°C T°final - 37620 J + 418 J/°C T°final - 4180J = 0

836 J/°C T°final - 37620J - 4180J = 0

836 J/°C T°final = 41800J

T°final =  41800J / 836 °C/J = 50°C

1.solid

2.carbon dioxide

3.helium

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 :