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Blababa [14]
2 years ago
15

The oxidation number of what atom or ion decreases in the following reaction?

Chemistry
1 answer:
Flauer [41]2 years ago
7 0

The oxidation number of I (iodine) decreases the in the described chemical reaction from 0 to -1.

Explanation:

We have the following chemical reaction:

I₂ + 2 KCl → KI + Cl₂

The following atoms have a change in the oxidation number:

I with a 0 oxidation number (in I₂) goes to I with -1 oxidation number (in KI) by receiving 1 electron.

Cl with a -1 oxidation number (in KCl) goes to I with 0 oxidation number (in Cl₂) by losing 1 electron.

Learn more about:

oxidation number

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What is the correct order of the steps in the scientific<br> method?
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3 years ago
How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
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Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

Now put all the given values in the above expression, we get:

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

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