A 34.0 g piece of metal is heated to 92.0°C then placed in a beaker of water containing 22.0 g of water at 19.0°C. The temperature of the water rises to 24.0°C. What is the specific heat of the metal?

0.1988 J/g°C

Explanation:

-Qmetal = Qwater

Q = mc∆T

Where;

Q = amount of heat

m = mass of substance

c = specific heat of substance

∆T = change in temperature

Hence;

-{mc∆T} of metal = {mc∆T} of water

From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.

For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?

Note that, the final temperature of water and the metal = 24°C

-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)

-{34 × c × (-68°C)} = 459.8

-{34 × c × -68} = 459.8

-{-2312c} = 459.8

+2312c = 459.8

c = 459.8/2312

c = 0.1988

The specific heat capacity of the metal is 0.1988 J/g°C

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24