This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.
1 mole is equal to 6.022x10²³ particles as given, so:

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Answer:</h3>
2.49 mol
Let me know if you have any questions.
The 2nd ionization energy is removing a 2nd electron from that resulting cation:
<span>Li+ --> Li2+ + 1e- </span>
The molecular weight of a given compound would simply the
sum of the molar weights of each component.
The molar masses of the elements are:
C = 12 amu
H = 1 amu
N = 14 amu
O = 16 amu
where 1 amu = 1 g / mol
Since there are 6 C, 5 H, 1 N and 2 O, therefore the
total molecular weight is:
molecular weight = 6 (12 amu) + 5 (1 amu) + 1 (14 amu) +
2 (16 amu)
molecular weight = 123 amu
Therefore the molecular weight of nitrobenzene is 123 amu
or which is exactly equivalent to 123 g / mol.
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.