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mr Goodwill [35]
3 years ago
14

How did the work of Dmitri Mendeleev differ from that of John Newlands in the development of the periodic table?

Chemistry
1 answer:
alex41 [277]3 years ago
3 0
When Newlands tried to create a periodic table, his tried to conform to the "Rule of Octaves" he had discovered. He had the right idea, in that if you arrange the elements by atomic weight there would be similarities every 7 elements (not 8 because noble gases hadn't been discovered yet) but he tried to push this rule so much that he would put multiple elements in the same box to try to keep the rule. Mendeleev, however, left gaps in this table for undiscovered elements, which paved the way for our modern periodic table.
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When the nail was copper-plated in Part III of the lab, what element do you think was reduced? Explain your answer in complete s
Jet001 [13]
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.
6 0
3 years ago
What does the nucleus of the cell control?
dusya [7]

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:

7 0
3 years ago
Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with
skad [1K]

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.

Learn more:

Source of energy brainly.com/question/2948717

#learnwithBrainly

4 0
3 years ago
A Carnot cycle operates between the temperatures limits of 400 K and 1600 K, and produces 3600 kW of net power. The rate of entr
TiliK225 [7]

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 = \frac{WT_{H}}{T_{H}- T_{L}}

The entropy of a system is a measure of how disorderly a system is getting. The rate of entropy generation during heat addition is,

S_{gen} = \frac{Q}{T_{H}} = \frac{W}{T_{H} - T_{L}}

Calculation:

<u>Given:</u>

T_{L} = 400K

T_{H} = 1600K

W = 3600 kW

Put all the values in the above equation, and we get,

S_{gen} = \frac{W}{T_{H} - T_{L}} = \frac{3600}{1600-400} = 3 kW/K

The rate of entropy change is 3 kW/K

Learn more about the Carnot cycle here,

brainly.com/question/13002075

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3 0
2 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
Kay [80]

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

3 0
3 years ago
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