Reduction <span>always results in a lowering of the oxidation number. The reaction of the system above is written as:
</span><span>Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq)
</span>
From the reaction, we see that copper goes from the +2 to a neutral charge. Lowering of the oxidation number happens so this is the element that is being reduced.
Answer:
Image result for What does the nucleus of the cell control?
The nucleus controls and regulates the activities of the cell (e.g., growth and metabolism) and carries the genes, structures that contain the hereditary information. Nucleoli are small bodies often seen within the nucleus.
Explanation:
A. Decomposing water requires a high activation energy.
Explanation:
In decomposing water to release hydrogen gas to make fuel cells, the process requires a very high activation energy.
2H₂O ⇆ 2H₂ + O₂
This is the overall reaction. O-H must be broken to release free hydrogen to produce hydrogen gas.
The O-H bond is a very strong force of attraction that requires a high activation energy to overcome.
- The activation energy is the energy barrier that must be overcome before a reaction takes place.
- The sun is a renewable source of energy.
- Water decomposition produces useful oxygen gas needed by all life for cellular respiration.
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The rate of entropy change:
The rate of entropy change of the working fluid during the heat addition process is 3 kW/K
What is the Carnot cycle?
- The Carnot Cycle is a thermodynamic cycle made up of reversible isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression processes in succession.
- The ratio of the heat absorbed to the temperature at which the heat was absorbed determines the change in entropy.
The entropy of a system:
The rate of heat addition is expressed as,
Q = 
The entropy of a system is a measure of how disorderly a system is getting. The rate of entropy generation during heat addition is,

Calculation:
<u>Given:</u>
= 400K
= 1600K
W = 3600 kW
Put all the values in the above equation, and we get,
=
= 3 kW/K
The rate of entropy change is 3 kW/K
Learn more about the Carnot cycle here,
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Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :

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})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%5D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene = 
= specific heat of liquid benzene = 
= enthalpy change for fusion = 
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]](https://tex.z-dn.net/?f=Q%3D%5B94.4g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28279-322%29K%5D%2B%5B94.4g%5Ctimes%20-125.6J%2Fg%5D%2B%5B94.4g%5Ctimes%201.51J%2Fg.K%5Ctimes%20%28205-279%29K%5D)

Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ