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Vlada [557]
3 years ago
7

What is the oxidation state of an individual sulfur atom in so42−?

Chemistry
1 answer:
hichkok12 [17]3 years ago
4 0

The oxidation state of single oxygen atom is -2 since it is more electronegative and like to gain electrons.

The sum of the oxidation states of each atom = charge of compound

Let’s say the oxidation state of Sulfur is x.

Hence,

<span>      x +4 * (-2) = -2</span>

<span>      x – 8         = -2</span>

<span>                  x   = +6</span>

<span>The oxidation state of individual Sulfur atom is +6</span>

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Answer:

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Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g

Best regards.

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