Ionic bonds are formed when there is complete transfer of valence electrons between two atoms.
Electronegativity tells the trend of an atom to atract electrons.
You should search for the complete set of rules that indicate whether an ionic or covalent bond happens.
There are two relevant rules to state if whether an ionic bond will happen:
- When the difference of electronegativities between the two atoms is greater than 2.0, then the bond is ionic.
- When the difference is between 1.6 and 2.0, the bond is ionic if one of the elements is a metal.
You need to list the electronegativities of the five elements (there are tables with this information)
Element electronegativity
Cu: 1.9
H: 2.2
Cl 3.16
I: 2.66
S: 2.58
Differences:
Cu / S: 2.58 - 1.9 = 0.68
H / S: 2.58 - 2.2 = 0.38
Cl / S: 3.16 - 2.58 =0.58
I / S: 2.66 - 2.58 = 0.08
Those differences are too low to consider that the bond is ionic.
Then the answer is that none of those atoms forms an ionic bond with sulfur.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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