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3 years ago

How does particles of substance behave at its melting point?

2 answers:

7 0

Answer: The particles of the substace move away from each other and slip breaking the crystalline structure.


Explanation:


Melting is the change of phase from solid to liquid.


Melting is the result of adding heat to the solid.


The particles in a solid are arranged in fixed structures, called lattices, in which the partilces are packed closed together to each other, and so they cannot move one respect to others.


At the melting point, as heated is added, the partilcles vibrate and move more rapidly, gaining kinetic energy.


The kinetic energy gained by the heat transferred is not used to increase the temperature but to increase the speed of the particles which finally manage to slip out of the fixed structure and so become liquid.

ivolga24 [154]3 years ago

5 0

When a substance is heated, its atoms gain energy and begin to vibrate rapidly within the lattice of the substance (the substance expands). As more heating continues, the atoms gain even more energy and move more rapidly until they are able to overcome the force of the bonds, that hold the atoms together in the lattice, until the structure is disrupted.

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⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

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