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Ludmilka [50]
1 year ago
12

What is the source of energy in nuclear power plants?

Chemistry
1 answer:
Zinaida [17]1 year ago
4 0

Answer:

<h2><em><u>URANIUM</u></em><em><u> </u></em></h2>

Explanation:

What is the source of energy in nuclear power plants?

<em><u>Uranium</u></em> is the fuel most widely used by nuclear plants for nuclear fission. <u>Uranium</u> is considered a nonrenewable energy source, even though it is a common metal found in rocks worldwide. Nuclear power plants use a certain kind of uranium, referred to as U-235, for fuel because its atoms are easily split apart.

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The molar volume of oxygen,O2, is 3.90 dm3 mol-1 at 10.0 bar and 200 degree centigrade. Assuming that the expansion may be trunc
schepotkina [342]

Answer:

B = - 0.0326 dm³/mol

Explanation:

virial eq until second term:

  • PVm = RT [ 1 + B/Vm ]

∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm

∴ T = 200°C = 473 K

∴ Vm = 3.90 dm³/mol

∴ R = 0.08206 dm³.atm/K.mol

⇒ PVm / RT = 1 + B/Vm

⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm

⇒ 0.99164 = 1 + B/Vm

⇒ B/Vm = - 8.357 E-3

⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )

⇒ B = - 0.0326 dm³/mol

4 0
3 years ago
Identify the hybridization of each carbon atom for the molecule above
ipn [44]

Carbons starting from the left end:

  1. sp²
  2. sp²
  3. sp²
  4. sp
  5. sp

Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

  • A carbon atom with four electron domains is sp³ hybridized;
  • A carbon atom with three electron domains is sp² hybridized;
  • A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

  • The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
  • The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
  • The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
  • The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
  • The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.

8 0
3 years ago
The molecule CH20 contains two single bonds and one double bond.<br> True<br> False
dem82 [27]

Answer:

True

Explanation:

The molecule CH20 contains two single bonds and one double bond.

6 0
3 years ago
For the balanced chemical reaction
musickatia [10]

Answer:

150

Explanation:

  • C₄H₂OH + 6O2 → 4CO2 + 5H₂O

We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:

  • 100 molecules CO₂ * \frac{6moleculesO_2}{4moleculesCO_2} = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

5 0
3 years ago
What is the edge length of a face-centered cubic unit cell that is made of of atoms, each with a radius of 154 pm
gladu [14]

Answer:

The edge length of a face-centered cubic unit cell is 435.6 pm.

Explanation:

In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.

Hence, the number of atoms in an FCC unit cell is:

8*\frac{1}{8} + 6*\frac{1}{2} = 4 atoms

In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:

a^{2} + a^{2} = (4R)^{2}     (1)

Where:

a: is the edge length

R: is the radius of each atom = 154 pm      

By solving equation (1) for "a" we have:

a = 2R\sqrt{2} = 2*154 pm*\sqrt{2} = 435.6 pm    

Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.   

I hope it helps you!

7 0
3 years ago
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