14. In the lab, an experimenter mixes 75.0 g of water (initially at 30oC) with 83.8 g of a solid metal (initially at 600oC). At thermal equilibrium, he measures a final temperature of 50.0oC. What metal did the experimenter probably use

Tungsten is used for this experiment

Explanation:

This is a Thermal - equilibrium situation. we can use the equation.

Loss of Heat of the Metal = Gain of Heat by the Water

                    Q = mΔT

Q = heat

m = mass

ΔT = T₂ - T₁

T₂ = final temperature

T₁ = Initial temperature

Cp = Specific heat capacity

Metal

m = 83.8 g

T₂ = 50⁰C

T₁ = 600⁰C

Cp =

Water

m = 75 g

T₂ = 50⁰C

T₁ = 30⁰C

Cp = 4.184 j.g⁻¹.⁰c⁻¹

⇒ - 83.8 x x (50 - 600) = 75 x 4.184 x (50 - 30)

⇒ = j.g⁻¹.⁰c⁻¹

We know specific heat capacity of Tungsten = 0.134 j.g⁻¹.⁰c⁻¹

So metal Tungsten used in this experiment

imma come back to you hold on

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

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