More Problems! Are There NO AP Chem people out there?!? Somebody has to know some of these!!! I need HELP!!!

Settle down... You can do this.

#1 You have a book and class notes, yes?

I will use c to mean specific heat:

-heat lost by metal = heat gained by water

-mc∆T for the metal = mc∆T for the water

- (25.0g)c(24.4°C - 88.0°C) = (150.0g)(4.184 J/g°C)(4.4°C)

Solve for c -- specific heat of metal

#2 I assume you have typos on second line down -- should be C2H2, right? And you can't have SO2 as reactants in both equations without a sulfur compound as products. So something else going on there. Do you mean 5 CO2? And 2H2O? (Your equation isn't balanced.) After you've fixed all that, you are going to use Hess's Law to solve this problem. That is, arrange the two equations w/ ∆H so that they can add up to the equation you want; then you can add the two ∆H's to get the ∆H you want.

You can flip equations, but change the sign of ∆H; you can multiply the equation by a number (or fraction), but multiply ∆H by the same number.

1st eq: Flip and multiply by 3/2:

6CO2 + 3H2O --> 3 C2H2 + 15/2 O2 ∆H = 3/2 x (+2600kJ)

2nd eq: Multiply equation and ∆H by 1/2; add first and second eq. It should equal the equation you wanted. Now the ∆H's will add up the the ∆H of the equation you wanted.

#3 For this one you have to know that enthalpy of formation is the energy change if you make one mole from free elements.

1/2 N2 + 1/2 O2 --> NO ∆Hf° = +90.3kJ/mol

1/2 N2 + O2 --> NO2 ∆Hf° = +33.1 kJ/mol

Flip and multiply x 2 the 1st eq. Just x 2 the 2nd. Add these two to get the equation

2NO + O2 --> 2NO2

∆H = 2(-90.3) + 2(+33.1)

Good luck!