A piece of metal weighing 10.0 g is raised in temp. from, 25 C to 76.5 C. it\'s

Question

A piece of metal weighing 10.0 g is raised in temp. from, 25 C to 76.5 C. it's found that it absorbed 47.6 calories of heat in the process. Calculate the spesific heat (S.H.) of the metal. Show your work and include the proper units

1) idebtify the metal in the example above from this list

Element: Spesific heat (cal/gC-degree)
Aluminum 0.2158
Copper 0.0928
iron 0.1070
Silver 0.0566

Identify__________

2) Determine the enthaply change ( "delta H" ) of the reaction indicated?

S(s) + O2 ---> SO2(g) (delta H) = -471 Kacl
+ SO2(g) + fi O2(g) ---> SO3(g) (delta H) = -22 Kacl
_______________________________________________
S(s) + 3/2 O2 ---> SO3(g) (delta H) = ???


3) What would be the enthalpy change for the reverse of the last reaction above?

please I need help with these questions above, thanks

Explanation / Answer

(1) Specific heat = heat absorbed/(mass x temperature change)

= 47.6/(10.0 x (76.5 - 25)

= 0.0924 cal/g.deg C

Identity of metal (with closest specific heat value): Copper (0.0928 cal/g.deg C)

(2) Since reaction indicated is obtained by the sum of the other two reactions:

Delta H(forward) = Delta H(1) + Delta H(2)

= -471 + (-22)

= -493 kcal

(3) Delta H(reverse) = -Delta H(foward)

= +493 kcal

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