<u>Answer:</u> The given compound is a covalent compound.
<u>Explanation:</u>
Covalent compound is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound. These are usually formed when two non-metals react.
An ionic compound is defined as the compound which is formed when electron gets transferred from one atom to another atom. These are usually formed when a metal reacts with a non-metal.
We are given:
A chemical compound having chemical name as nitrogen triiodide.
This compound is formed by the combination of nitrogen and iodine atoms. Both these elements are non-metals and thus form covalent compound.
The chemical formula for the given compound is
Hence, the given compound is a covalent compound.
Answer:
0.38
Explanation:
As you go from 0 % hexane (χ = 0) to 100 % hexane (χ = 1), the refractive index decreases from 1.407 to 1.375, a decrease of 0.032.
From 1.407 to 1.3948, the decrease is 0.0122 .
That is 0.0122/0.032 = 0.38 of the distance from χ = 0 to χ = 1.
The mole fraction of hexane is 0.38.
The diagram below is a plot of refractive index vs. the mole fraction of hexane. It shows that the refractive index drops to 1.3948 when χ = 0.38.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.