Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles = 
C; molar mass = 12;
Number of moles = 
= 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles = 
= 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce 
= 1.79moles of CF₄
<span>If 36 gm of potassium chlorate enter into the reaction, the total mass of the two products will still be 36 gm because if there is only one reactant, the mass of the compounds after the reaction will be same that reactant based on the law of conservation of matter.</span>
<u>Ionic Bond</u> is formed when the electronegativity difference is 0.4 > 2.0. Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.
Explanation:
Electronegativity is a term that can be defined as a tendency of an atom to attract electron towards its own self.
An electronegativity of an atom is affected by
- The atomic number of the atom
- Secondly by the distance at which the valence electron are residing from the nucleus
1. In case the electronegativity difference (which is denoted by ΔEN) is less than 0.5 then the bond formed is known as N<u>onpolar covalent.
</u>
2. In case the ΔEN is in between 0.5 and 1.6, the bond formed is referred to as the<u> Polar covalent
</u>
3. In case the ΔEN is more /greater than 2.0, then the bond formed is referred to as<u> Ionic Bond</u>
<u>2 Examples of Ionic bonds</u>
- The formation of sodium fluoride, NaF, from a sodium atom and a fluorine atom is an example of Ionic bond formation.
- Another example is the formation of NaCl from sodium (Na),which is a metal, and chloride (Cl), which is a nonmetal
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
Answer:
According to Bohr, the amount of energy needed to move an electron from one zone to another is a fixed, finite amount. ... The electron with its extra packet of energy becomes excited, and promptly moves out of its lower energy level and takes up a position in a higher energy level. This situation is unstable, however.
