110 g of water (specific heat = 4.184 j/g c) and 100 g of a metal sample (specific heat = 0.397 j/g c) are heated from 25 degrees c to 75 degrees c. which substance required more thermal energy?

Water.

Explanation:

Q = m.C.ΔT, where,

Q is the thermal energy,

m is the mass of the substance ,

C is the specific heat of the substance ,

ΔT is the increase in temperature.

For water,

Q for water = m.C.ΔT = (110 g) (4.184 J/g.°C) (75 °C - 25 °C) = 23012 J.

For the substance,

Q for the substance = m.C.ΔT = (100 g) (0.397 J/g.°C) (75 °C - 25 °C) = 1985 J.

So, it is clear that water (23012 J) requires more thermal energy than the substance (1985 J).

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 :

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