2. a 237 g piece of molybdenum, initially at 100.0 °c, was dropped into 244 g of
water at 10.0°c. when the system came to thermal equilibrium, the temperature
was 15.3°c. what is the specific heat capacity of molybdenum? the specific heat
capacity of water is 4.184 j/g. k. (2.5 points)

Explanation:

1. Energy balance

By the first law of thermodynamics, considering the system is closed and isolated, the heat released by the 237 g piece of molybdenum equals the heat absorbed by the 244 g of water.

2. Heat equation

The heat released or absorbed by a substance is proportional to the product of the mass, the specific heat and the change in temperature

3. Heat released by the 237 g piece of molybdenum

At equilibrium:

4. Heat absorbed by the 244 g of water

At equilibrium:

5. Solve for C from Q₁ = Q₂

The result must be reported with 3 significant figures: C = 0.270 J/gºC.

i believe it is-

d.) gravity pulled particles of dust and gas together to form planets.

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 :