Hepta is seven and tetra is four, so option c is your answer
Iron III Chloride has a chemical formula of FeCl₃, while ammonium hydroxide has a chemical formula of NH₄OH.
The <em>balanced equation</em> would be:
FeCl₃ (aq) + 3 NH₄OH (aq) → Fe(OH)₃ (s) + 3 NH₄Cl (aq)
The precipitate is Fe(OH)₃ or iron iii hydroxide.
To find the <em>complete ionic equation</em>, dissociate the compounds in aqueous phases into their ionic forms:
Fe³⁺ + Cl⁻ + NH₄⁺ + 3 OH⁻ --> Fe(OH)₃(s) + NH₄⁺ + Cl⁻
To find the <em>net ionic equation</em>, cancel out like ions that appear both in the reactant and product side:
Fe³⁺ + 3 OH⁻ --> Fe(OH)₃
The biological risk for the first person than the second as a result of radiation weighting is 10 times.
<h3>
What is radiation weighting factor?</h3>
As stated in the question, radiation weighting factor (q) is the ability to transfer energy to the body.
If radiation factor of proton = 2, and radiation factor of alpha particles = 20.
- First person is exposed to alpha radiation = 20
- Second person is exposed to protons = 2
Risk of first person with respect to second person = 20/2 = 10 times higher
Learn more about radiation factor here: brainly.com/question/24039736
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Answer:
A cation has more protons than electrons, consequently giving it a net positive charge. For a cation to form, one or more electrons must be lost, typically pulled away by atoms with a stronger affinity for them.
Were i found my answer: Cation vs Anion: Definition, Chart and the Periodic Table
Explanation:
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
