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hjlf
1 year ago
15

Fusion and fission reactions are both nuclear reactions that can be used to produce energy. However, while fission reactions are

observed as a natural decay route for some nuclei on Earth, fusion is not seen under typical ambient planetary conditions. More extreme conditions, like are present in stars, are typically necessary for fusion to occur on a large scale.1. Which of the answers below correctly describes the reasoning for this difference?A) Both fusion and fission reactions are initiated by neutron addition to the nuclei involved in the reaction. However, fusion reactions, unlike fission reactions, do not produce large amounts of extra neutrons to propagate the chain reaction needed to sustain a reaction at a high rate.B) Both fusion and fission reactions require nuclear collisions. Fusion reactions involve smaller nuclei which collide less frequently with each other, causing a slower reaction. Fission reactions involve larger nuclei, making collisions more frequent, leading to faster reaction rates.C) Fusion reactions result in less energy released as heat than is seen in fission reactions. Because less heat is released there is less energy present in the reactants to overcome the large activation energy for these reactions. Fission reactions also have a large activation barrier, but the heat produced by these reactions is much greater, giving the nuclei the energy they need to complete the reaction.D) Fusion reactions have a larger barrier to reaction due to the repulsion forces required for two nuclei to come together. Because the nuclei are both positively charged, the repulsive force between the two has to be overcome for fusion to occur. Fission reactions do not involve nuclear collisions and therefore have a lower barrier to reaction.E. The binding energy per nucleon is much lower, on average, for atoms involved in fusion reactions. This means that the reactions are less energetically favorable, as binding energy per nucleon is a measure of stability. Because there is much less energetic driving force, the reaction rate is much slower overall, meaning that it is more difficult to observe under ambient conditions.
Chemistry
1 answer:
kipiarov [429]1 year ago
3 0

Answer:

I dont say bla bla bla

Explanation:

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If 1.00 mol of argon is placed in a 0.500-L container at 19.0 ∘C , what is the difference between the ideal pressure (as predict
lana66690 [7]

41.083 atm is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation.

Explanation:

Data given for argon gas:

number of moles = 1 mole

volume = 0.5 L

Temperature = 19 degrees or 292.15 K

a= 1.345 (L2⋅atm)/mol2

b= 0.03219L/mol.

R = 0.0821

The real pressure equation given by Van der Waals equation:

P =( RT ÷ Vm-b) - a ÷ Vm^2

Putting the values in the equation:

P = (0.0821 x 292.15) ÷(0.5 - 0.03219) - 1.345÷ (0.5)^2

  = 23.98÷0.4678 - 1.345 ÷0 .25

  = 51.26 - 5.38

  = 45.88 atm is the real pressure.

The pressure from the ideal gas law

PV =nRT

P =( 1 x 0.0821 x 292.15) ÷ 0.5

  = 4.797 atm

the difference between the ideal pressure and real pressure is

Pressure by vander waal equation- Pressure by ideal gas law

45.88 - 4.797

= 41.083 atm.is the difference between the two.

4 0
1 year ago
Identify the items in the list below that are examples of matter. Check all that apply. idea iron sugar sound glass neon gas Cho
Harlamova29_29 [7]
Sugar is a solid. glass is a solid. gas is well gas
5 0
2 years ago
Read 3 more answers
Which of the following shows the correct rearrangement of the the heat equation q = mCpΔT to solve for specific heat?
strojnjashka [21]
When 
q = m*Cp*ΔT

when q is Heat energy in Joules

and m is the mass of the substance in Kg

and Cp is the specific heat (J/Kg.K)

and Δ T is the change in temperature in Kelvin


so, by rearranging the formula we can get the specific heat Cp from:

∴Cp = q / m*ΔT
6 0
2 years ago
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Please help me double-check my answer: Calculate the molarity of an aqueous solution that contains 36.5g KMnO4 and has a total v
Helen [10]

Answer:

The answer to your question is Molarity = 0.6158, I got the same answer as you.

Explanation:

Data

Molarity = ?

Mass of KMnO₄ = 36.5 g

Total volume = 375 ml

Process

1.- Calculate the Molar mass of KMnO₄

KMnO₄ = (1 x 39.10) + (54.94 x 1) + (16 x 4)

            = 39.10 + 54.94 + 64

            = 158.04 g

2.- Calculate the moles of KMnO₄

                158.04 g of KMnO₄ ------------------- 1 mol

                  36.5 g of KMnO₄ ---------------------  x

                   x = (36.5 x 1) / 158.04

                   x = 0.231 mol

3.- Convert the volume to liters

                  1000 ml -------------------- 1 L

                    375 ml --------------------- x

                     x = (375 x 1)/1000

                    x = 0.375 L

4.- Calculate the Molarity

Molarity = moles / volume

-Substitution

Molarity = 0.231 moles / 0.375 L

Result

Molarity = 0.6158

6 0
2 years ago
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Define the following terms - you may need to consult your lecture text or other suitable resource:
natta225 [31]

Answer:

a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.

b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.

c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.

d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers.  The cross-links leads to an alteration in the physical properties of the polymer.

6 0
1 year ago
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