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Allisa [31]
3 years ago
11

What effect has impurities on the melting and boiling points of pure substances? illustrate your answer with an example

Chemistry
1 answer:
Andrei [34K]3 years ago
7 0

Answer:

The presence of an impurity in a substance lowers the melting point, and raises the boiling point of the substance. For example, Impurities present in the water decrease the freezing temperature of the water, and in the same way, impurities increase the boiling point of the water.

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The answer to this question is 159.609 g/mol
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Explain what is water of crystallization​
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In chemistry, water(s) of crystallization or water(s) of hydration are water molecules that are present inside [crystal]s. Water is often incorporated in the formation of crystals from aqueous solutions. ... Water of crystallization can generally be removed by heating a sample but the crystalline properties are often lost

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2 years ago
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
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Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

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~Na+1 is already in the preferred form. Because of this, the second ionization energy of sodium is higher than normal. Mg+1 loses an electron to form s2 p6 .

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