The statement that is NOT true of fission is B. slow neutrons are fired at a large nucleus.
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What is Nuclear fission</h3>
This is the splitting of large nuclei to smaller ones. It could be a natural or spontaneous dissociation by radioactive decay, or can be initiated in in a laboratory under required conditions. It involves;
- Splitting of large nuclei to two smaller ones
- Release of large amount of energy
- Neutrons released can be used to initiate chain reactions
Hence, the statement that is not true about fission is that slow neutrons are fired at a large nucleus.
Learn more about nuclear fission here:
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Answer:</h3>
= +0.43 V
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Explanation:</h3>
The two standard reduction potentials are;
Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E₀ = +0.77 V
Cu²⁺(aq) + 2e⁻ → Cu(s) E₀ = +0.34 V
The overall reaction is;
Fe³⁺(aq) + Cu(s) → Fe²⁺(aq) + Cu²⁺(aq)
To calculate the standard cell potential of the overall reaction, we use the equation;
Ecell = E redu, cathode - E red, anode
From the reaction; Copper lost electrons to form copper ions (Cu²⁺) (Oxidation) while Fe³⁺ gained electrons to form Fe²⁺ (reduction)
Therefore;
E cell = + 0.77 v - (+0.34 v)
= + 0.43 V
Hence, the standard cell potential of the overall reaction is +0.43 V
Answer:
Explanation:
A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.
This is an arrhenius Base
According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :
- Sodium hydroxide NaOH
- Potassium hydroxide KOH
A substance that produces an excess of hydrogen ion (H+) in aqueous solution
This is an arrhenius Acid
An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.
Examples are;
- Hydrochloric acid HCl
- Hydroiodic acid HI
- Hydrobromic acid HBr
Answer:
The concentration of the diluted solution is 3.75 M
Explanation:
Step 1: Data given
Volume of a stock solution = 75 mL = 0.075 L
The molarity of stock solution = 5.0 M
We add 25 mL, The total volume is 100 mL = 0.100 L
Step 2: Calculate the concentration of the diluted solution
C1*V1 = C2*V2
⇒with C1 = the initial concentration = 5.0 M
⇒with V1 = the initial volume = 0.075 L
⇒with C2 = the new concentration = TO BE DETERMINED
⇒with V2 = the diluted volume = 0.100 L
5.0 M * 0.075 L = C2 * 0.100L
C2 = (5.0M * 0.075 L) / 0.100 L
C2 = 3.75 M
The concentration of the diluted solution is 3.75 M
Answer:
Boron (B) is the element whose IE matches with our data.
Electronic Configuration of boron:
Explanation:
Ionization Energy (IE):
It is the minimum amount of energy which is required to remove the lose electron. If the electron is closer to the nucleus then greater amount of energy is required to remove the electron.
If we look from left to right in a period, ionization energy increases due stability of valance shell.
From the data given to us:
IE₁ = 801
IE₂ = 2427
IE₃ = 3659
IE₄ = 25,022
IE₅ = 32,822
Boron (B) is the element whose IE matches with our data.
Electronic Configuration of boron:
Boron has 5 electrons (3 in valance shell) that's why it has 5 Ionization Energies.