Answer:
1. Hydrogen Iodide
2. 6 molecules of Hydrogen Iodide
3. Iodine is the limiting reagent
Explanation:
The image of the illustration in the question has been attached:
1. The illustration represents a mixture of hydrogen ( light blue ) and iodine ( purple )
H₂ + I₂ ---> 2HI
This forms hydrogen iodide.
2. In the given illustration, 6 product molecules of Hydrogen Iodide. This is indicated in the box on the right side of the illustration.
3. The limiting reagent is the reactant that determines how much of the products are made. It is the substance that is totally consumed when the chemical reaction is completed. In the box on the right side of the illustration, you will see that hydrogen which is indicated by blue is in excess. The limiting reagent is the one that is completely consumed which is the iodine.
Answer:
<h3>A</h3>
Explanation:
<h2>sorry try lang baka mali</h2><h2 />
For moles to grams would be the mole which is 7.9*10^-1 times the molar mass of argon
This one is an oxidation-rdcution equation:
<span>H2 + F2 → 2HF
How can you tell?
If the oxidation states of the atoms in the reactans are different from the oxidation states of the same atoms in the products then it is an oxidation-reduction reaction.
Both atoms H and F in the reactants have oxidation states 0.
That is a basic rule: any atom alone or bonded to the same kind of atom has oxidation state 0.
The oxidation states in HF are: H: +1, and F: -1.
So, the H increased its oxidation state, which is that ii is oxydized ; while F reduced its oxidation state so it is reduced.
Answer: H2 + F2 ----> 2HF
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