Answer:
BF3
Explanation:
For this question, you need to use the number of valence electrons present in each element. Boron is in group 3/13 on the periodic table so you know it has 3 valence electrons while Fluorine is in group 7/17 so it has 7 valence electrons. These elements are both covalent so they will share electrons. All elements in the first three rows want to reach either have 8 valence electrons or zero valence electrons depending on whichever is easier. When B and F interact each Fluorine will only want to take one electron, but Boron wants to get rid of all 3 electrons, so it will bond with 3 Fluorine to get rid of all its valence electrons.
I hope this helps.
The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
Answer:
the concentration of PCl5 in the equilibrium mixture = 296.20M
Explanation:
The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.
The detaied steps is as shown in the attached file.
CH3 is the empirical formula for the compound.
A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.
The number of atom or moles in the compound is
1.17 g C X 1 mol of C / 12.011 g C = 0.097411 mol of C.
0.287 g H x 1 mol of H / 1 g H = 0.28474 mol H.
This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.
So we can represent the compound with the formula C0.974H0.284.
Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.
we can divide 0.284 by 0.0974.
0.284 / 0.0974 = 3.
So here, Carbon is one and hydrogen is 3.
We can write the above formula as a CH3.
Hence the empirical formula for the sample compound is CH3.
For a detailed study of the empirical formula refer given link brainly.com/question/13058832.
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