Suppose an Olympic diver who weighs 55.0 kg executes a straight dive from a 10-m

Question

Suppose an Olympic diver who weighs 55.0 kg executes a straight dive from a 10-m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water.

Part A

What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? Ep=

Part D

Suppose that 0.290 mol of methane, CH 4 (g) , is reacted with 0.440 mol of fluorine, F 2 (g) , forming CF 4 (g) and HF(g) as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A )

potential energy=mgh

Ep = 55*9.8*10.8= 5821.2 J

set the potential energy equal to kinetic energy:

D)  2CH4 + 5F2 --> 2CF4 + 2HF

the enthalapy of formation of the compounds are as table .
CH4 = -74.8kJ/mole
CF4 = -925.0 kJ/mole
HF = -271.1 kJ/mole

0.29 moles CH4 x (5F2 / 2CH4) = 0.725moles F2 required. we only have 0.44moles. F2 = limiting reactant
0.44moles F2 yields 2/5 x 0.44mole = 0.176moles CF4 and HF, requires 0.176moles CH4
deltaHrxn = products - reactants
(0.176 x -271.1kJ/mole + 0.176 x -925kJ/mole) - (0.176 x -74.8kJ/mole)
-19734.88 kJ

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