Note that it says oxygen "gas"
So you need the atomic mass of oxygen gas
Look at your periodic table, you'll see 15.9994 under oxygen
Oxygen gas has a formula of O2 therefore,
(15.9994) times 2= Oxygen gas atomic mass=31.9988
Mol= Mass/Atomic Mass
=62.3 g/ 31.9988 g/mol = 1.95 mol
now look at the ratio of C2H6 and O2, notice there is an invisible number beside each of them, at that "invisible number" is =1
1 C2H6 + 1 O2 -> products
this means that for 1 mol of C2H6, 1 mol of O2 has to react with it
Thus as we have 1.95 moles of O2, we need 1.95 moles of C2H6
Explanation:
A period 3 element is one of the chemical elements in the third row (or period) of the periodic table of the chemical elements. The periodic table is laid out in rows to illustrate recurring (periodic) trends in the chemical behaviour of the elements as their atomic number increases: a new row is begun when the periodic table skips a row and a chemical behaviour begins to repeat, meaning that elements with similar behaviour fall into the same vertical columns. The third period contains eight elements: sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon. The first two, sodium and magnesium, are members of the s-block of the periodic table, while the others are members of the p-block. All of the period 3 elements occur in nature and have at least one stable isotope.[1]
Answer:
The correct option is;
4 percent ionic, 96 percent covalent, 222 pm
Explanation:
The parameters given are;
Phosphorus:
Atomic radius = 109 pm
Covalent radius = 106 pm
Ionic radius = 212 pm
Electronegativity of phosphorus = 2.19
Selenium:
Atomic radius = 122 pm
Covalent radius = 116 pm
Ionic radius = 198 pm
Electronegativity of selenium= 2.55
The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;
Using Pauling's alternative electronegativity difference method, we have;
![\% \, Ionic \ Character = \left [18\times (\bigtriangleup E.N.)^{1.4} \right ] \%](https://tex.z-dn.net/?f=%5C%25%20%5C%2C%20Ionic%20%5C%20Character%20%3D%20%5Cleft%20%5B18%5Ctimes%20%28%5Cbigtriangleup%20E.N.%29%5E%7B1.4%7D%20%20%5Cright%20%5D%20%5C%25)
Where:
Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36
Therefore;
![\% \, Ionic \ Character = \left [18\times (0.36)^{1.4} \right ] \% = 4.3 \%](https://tex.z-dn.net/?f=%5C%25%20%5C%2C%20Ionic%20%5C%20Character%20%3D%20%5Cleft%20%5B18%5Ctimes%20%280.36%29%5E%7B1.4%7D%20%20%5Cright%20%5D%20%5C%25%20%3D%204.3%20%5C%25)
Hence the percentage ionic character = 4.3% ≈ 4%
the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%
The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;
The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.
The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.
