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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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$