Alead–tin alloy of composition 30 wt% sn–70 wt% pb is slowly heated from a temperature of 150°c (300°) at what temperature does the first liquid phase form? (b) what is the composition of this liquid phase? (c) at what temperature does complete melting of the alloy occur? (d) what is the composition of the last solid remaining prior to complete melting?

a) 231.9 °C

b) 100% Sn

c) 327.5 °C

d) 100% Pb

Explanation:

This is a mixture of two solids with different fusion point:

Given that Sn has a lower fusion temperature it will start to melt first at that temperature.

So the first liquid phase forms at 231.9 °C and because Pb starts melting at a higher temperature, that phase's composition will be 100% Sn.

The mixture will be completely melted when you are a the higher melting temperature of all components (in this case Pb), so it will all in liquid phase at 327.5 °C.

At that temperature all Sn was already in liquid state and, therefore, the last solid's composition will be 100% Pb.

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

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