Answer:
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
Explanation:
Nuclei higher than Bi - 92 naturally are radioactive.
In a transmutation reaction, a new element is produced from an existing one due to radioactivity.
Nuclear fission is the radioactive process by which a heavy nucleus spontaneously decays into lighter ones with the release of a large amount of energy.
One example is the transmutation of uranium into thorium;
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
One chemical reaction is called the Haber process, a method for preparing ammonia by reacting nitrogen gas with hydrogen gas:
This equation shows you what happens in the reaction, but it doesn’t show you how much of each element you need to produce the ammonia. To find out how much of each element you need, you have to balance the equation — make sure that the number of atoms on the left side of the equation equals the number of atoms on the right.
You know the reactants and the product for this reaction, and you can’t change them. You can’t change the compounds, and you can’t change the subscripts, because that would change the compounds.
So the only thing you can do to balance the equation is add coefficients, whole numbers in front of the compounds or elements in the equation. Coefficients tell you how many atoms or molecules you have.
For example, if you write the following, it means you have two water molecules:
Each water molecule is composed of two hydrogen atoms and one oxygen atom. So with two water molecules (represented above), you have a total of 4 hydrogen atoms and 2 oxygen atoms.
You can balance equations by using a method called balancing by inspection. You take each atom in turn and balance it by adding appropriate coefficients to one side or the other.
With that in mind, take another look at the equation for preparing ammonia: HOPE THIS HELPS
<u>Given:</u>
Concentration of Cr2+ = 0.892 M
Concentration of Fe2+ = 0.0150 M
<u>To determine:</u>
The cell potential, Ecell
<u>Explanation:</u>
The half cell reactions for the given cell are:
Anode: Oxidation
Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V
Cathode: Reduction
Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V
------------------------------------------
Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)
E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V
The cell potential can be deduced from the Nernst equation as follows:
Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]
Here, n = number of electrons = 2
Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V
Ans: The cell potential is 0.418 V
Answer:
to help explore seas around them.
Explanation:
