A calorimeter contains 72.0 g of water at 19.2 oC. A 141 g piece of metal is heated to 89.0 oC and dropped into the water. The entire system eventually reaches 25.5 oC . What is the specific heat of the metal?

The specific heat of the metal is 0.212 J/(g°C).

Explanation:

We can calculate the specific heat of the metal by the following equilibrium:

In the above equation, we have that the heat loses by the metal (b) is gained by the water (a).

: is the water's mass = 72.0 g

: is the specific heat of water = 4.184 J/(g°C)            

: is the initial temperature of the water = 19.2 °C

: is the final temperature of the water = 25.5 °C

: is the metal's mass = 141 g

: is the specific heat of metal =?

: is the initial temperature of the metal = 89.0 °C

: is the final temperature of the water = 25.5 °C

Therefore, the specific heat of the metal is 0.212 J/(g°C).

I hope it helps you!

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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