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qwelly [4]
3 years ago
11

Elaborate on the difference in natural occurrences between fission and fusion reactions.

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
5 0
Fission is the division of one atom into two, and fusion is the combination of two lighter atoms into larger one. They are opposing processes and therefore very different.

Hoping that this is what you are looking for.
You might be interested in
What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0
3 years ago
1. A mixture contains 8.00 g each of O2, CO2, and SO2 at STP. Calculate the volume of this mixture. Which of the gases would exe
Gekata [30.6K]

Answer:

Explanation:

mole of O₂ = \frac{8}{32}

= .25 moles

mole of CO₂

= \frac{8}{44}

= .1818 moles

moles of SO₂

\frac{8}{64}

= .125 moles

Total moles of gas

= .5568 moles.

total volume of gas mixture

= 22.4 x .5568 liter ( volume of one mole of any gas = 22.4 liter)

= 12.47 liter.

gas will exert partial pressure according to their mole fraction

gas having greatest no of moles in the total mole will have greatest mole fraction so

O₂ will have greatest partial pressure.

7 0
3 years ago
9. Which series shows levels of organization in order from the smallest to the largest
erastova [34]

Answer:

B. Atom, molecule, cell, organism

Explanation:

Atoms make up molecules. Molecules make up cells. Cells make up tissue. A number of tissues working together creates an organ. Multiple organs working together creates an organ system. Organ systems working together creates one happy, functioning organism. :)

8 0
3 years ago
How many moles are in 25 g KMnO4
Ronch [10]

Answer:

0.158 moles KMnO4

Explanation:

According to the Periodic Table,

K = 39.10 g/mol

Mn = 54.94 g/mol

O = 16.00 g/mol

KMnO4 = 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol

25.0 grams KMnO4              1 mole

-----------------------------  x --------------------------  = 0.158 moles KMnO4

                                          158.04 grams

8 0
2 years ago
Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.
vladimir1956 [14]

<u>Given:</u>

Concentration of Ba(OH)2 = 0.348 M

<u>To determine:</u>

pOH of the above solution

<u>Explanation:</u>

Based on the stoichiometry-

1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion

Therefore, concentration of OH- ion = 2*0.348 = 0.696 M

pOH = -log[OH-] = - log[0.696] = 0.157

Ans: pOH of 0.348M Ba(OH)2 is 0.157

6 0
3 years ago
Read 2 more answers
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