1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
pychu [463]
3 years ago
8

How does nuclear fission of Uranium - 234 result in electricity being generated?

Chemistry
1 answer:
nevsk [136]3 years ago
7 0

Answer:

See explanation

Explanation:

The use of Uranium - 234 to generate electricity depends on a fission reaction. The uranium nuclide is bombarded by fast moving neutrons leading to a chain reaction. Control rods and moderators are used to keep the nuclear reaction under control.

As the nuclear reaction proceeds, heat is generated and steam is consequently produced. This steam is used to turn a turbine and electricity is thereby generated.

You might be interested in
Explain why a magnesium atom is smaller than atoms of both sodium and calcium.
hodyreva [135]
In the periodic table, there is a size trend.
The trend occurs horizontally and vertically on the table.
- From right to left, the atomic radius gets larger.
- From top to bottom, the atomic radius gets larger.

The reasoning for this is the additional energy levels and electrons being added as you go from right to left or top to bottom.
We can see that Magnesium is higher and more right in the periodic table than sodium and calcium.
Thus, Magnesium has a smaller atomic radius than both sodium and calcium
5 0
3 years ago
In general, ionization energies increase across a period from left to right. Explain why the second ionization energy of Cr is h
rodikova [14]

Answer:So this leads to the fact that second ionization energy  of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d  orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

Explanation:

Ionization energy is the amount of energy that we require to remove an electron form an isolated gaseous atom.

As we move from left to right across a period electrons are added to the same outermost shell therefore the attraction between the electrons and nucleus increases since more number of negatively charged electron are attracted to the positively charged nucleus.  This attraction leads to the decrease in atomic radii across a period and increase in ionization energy .

The increase in ionization energy occurs due to the fact that as the attraction  between the nucleus and outermost electrons increases so the electrons are more tightly bound to the nucleus hence more amount of energy is required to ionize the electron which leads to increase in ionization energy.

The electronic configuration of Cr and Mn are:

Cr:[Ar]3d⁵4S¹

Mn:[Ar]3d⁵4S²

The electronic configuration of Cr and Mn after 1st ionization:

Cr:[Ar]3d⁵4S⁰

Mn:[Ar]3d⁵4S¹

The electronic configuration of Cr and Mn after 2nd ionization:

Cr:[Ar]3d⁴4S⁰

Mn:[Ar]3d⁵4S⁰

As we can see that that 3d orbital of Cr (Chromium) is half filled with 5 electrons in it  and 4s orbital of Cr is also half-filled.

So when Cr is ionized for the first time then the electron from the half-filled 4s orbital will be removed .As the 1 electron present in outer most 4s orbital is removed so the 4s orbital now is completely vacant.

Now for the second ionization energy an electron ahs to be removed from half-filled 3d⁵ orbital. Hunds rule of maximum multiplicity states that the fully-filled or half-filled orbitals have maximum stability on account of symmetry and exchange energy.

So half-filled 3d⁵ orbital of Cr is very stable and hence to remove an electron from this would be require a lot of energy and hence the second ionization energy of chromium is higher than that of Manganese.

In case of Mn  the 3d orbital is also half -filled as chromium but the 4s orbital contains two electrons. when we remove the first electron from this orbital then also there is 1 electron present in the 4s orbital . So for the second ionization of Mn the only electron left in 4s orbital will be removed as the removal of electron from a 4s orbital is much easier as it requires less amount of energy as compared to  removal of  a electron from stable half filled 3d orbital.

So this leads to the fact that second ionization energy  of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d  orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

3 0
3 years ago
Identify one disadvantage to each of the following models of electron configuration:
Murrr4er [49]

Answer:

The disadvantages of each of the given model of electron configuration have been mentioned below:

1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.

2). Arrow and line diagrams make the counting of electrons and take up too much space.

3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.

6 0
3 years ago
Read 2 more answers
The pH of a 0.016-M aqueous solution of p-toluidine<br> (CH3C6H4NH2) is 8.60. Calculate Kb.
harkovskaia [24]

From the calculations performed, the Kb of the reaction is 9.9 * 10^-10

<h3>What is Kb?</h3>

The term Kb refers to the base dissociation constant of a solution. We have the following information from the question;

pH of the solution = 8.60

concentration of p-toluidine =  0.016-M

Recall that  pOH = 14 - pH

pOH = 14 - 8.60 = 5.4

[OH] = Antilog (-5.4) = 3.98 * 10^-6 M

We have to set up the ICE table;

          CH3C6H4NH2(aq) + H2O(l) ⇄ CH3C6H4NH3^+(aq) + OH^-(aq)

I             0.016                                           0                                   0

C           -x                                                   +x                                  +x

E    0.016  - x                                            3.98 * 10^-6                3.98 * 10^-6

Hence;

Kb = [3.98 * 10^-6]^2/[0.016  - 3.98 * 10^-6]

Kb =  9.9 * 10^-10

Learn more about Kb: brainly.com/question/12726841

8 0
2 years ago
2.0 mol of gas occupy 44.8 l at a temperature of 0°c. what is the pressure of the gas
MArishka [77]

The pressure of the gas is 1.0 bar.

<em>pV</em> = <em>nRT</em>

<em>T</em> = (0 + 273.15) K = 273.15 K

<em>p</em> = (<em>nRT</em>)/<em>V</em> = (2.0 mol × 0.083 14 bar·L·K⁻¹mol⁻¹ × 273.15 K)/44.8 L = 1.0 bar

5 0
4 years ago
Other questions:
  • 3. Which of these is a function of the<br> blood?
    10·1 answer
  • The florida everglades are one of the most
    5·1 answer
  • Blowing carbon dioxide gas into a solution of magnesium oxide will form what product?
    8·1 answer
  • The ionization energy of an element is
    11·1 answer
  • How do the rates of the forward and reverse reactions change throughout a reaction?
    11·1 answer
  • How many atoms of Titanium are in 3 moles?
    9·1 answer
  • What is the molarity of a solution of 1/2 mol of cu(no3)2 in 1/2 L water
    6·2 answers
  • Some transport processes use transport proteins in the plasma membrane, but do not require atp. this type of transport is known
    8·1 answer
  • Copper has two isotopes. Copper-63, which has an atomic mass of of 62.93 u and copper-65, which has an atomic mass of 64.93 u. I
    10·1 answer
  • What is the name of the molecular covalent compound kbr.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!