Help with Chemistry! Reaaaaaalllyyyyyyy Stuck.?

Question

1.Calculate the standard molar enthalpy of formation for sulfur dioxide.

S(s) + O2 (g)---> SO2 (g)
Given the thermochemical equations below.

2S (s) + 3 O2 (g)-------> 2 SO3 (g)
enthalpy change= -791.5 kJ

2SO2 (g) + O2 (g) ---> 2SO3 (g)
enthalpy change= -197.9 kJ

2. The specific heat capacity of aluminum is 0.897 J/g.K. What quantity of heat (in joules) is required to heat 15.0 g of Al from 22.0 degrees celsius to 75.0 degrees celsius?

3. Determine the heat of vaporization of Titanium(IV) chloride given the enthalpiesof reaction below.

Ti(s) + 2 Cl2 (s)---> TiCl4 (l) enthalpy change= -804.2 kJ

Ti(s) + 2 Cl2 (s)---> TiCl4 (g) enthalpy change= -763.2 kJ

Answer 1

1)
Use Hess's Law of Heat Summation:

2S (s) + 3 O2 (g) -------> 2 SO3 (g) [1st eqn.]
2SO3 (g) --------> 2SO2 (g) + O2 (g) [2nd eqn. reversed]
----------------------------------------------------------------------------------
2S (s) + 2O2 (g) --------> 2SO2 (g)

Standard enthalpy = (-791.5 + 197.9)/2 = -296.8 kJ

2)
q = c*m*dT

q = (0.897)(15.0)(348-295)

q = 713 J

3)
Vaporization is the change from liquid to gas. Manipulate your equations so that, when you add them, TiCl4 (l) is on the left, and TiCl4 (g) is on the right.

TiCl4 (l) ----> Ti (s) + 2Cl2 (s) [1st eqn. reversed]

Ti (s) + 2Cl2 (s) ----> TiCl4 (g) [2nd eqn.]

Heat of vaporization = 804.2 - 763.2 = 41.0 kJ

Answer 2

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