Wow very difficult I am so confused
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
Answer:
c. Compound 2 is more acidic because its conjugate base is more resonance stabilized
Explanation:
You haven't told us what the compounds are, so let's assume that the formula of Compound 1 is HCOCH₂OH and that of Compound 2 is CH₃COOH.
The conjugate base of 2 is CH₃COO⁻. It has two important resonance contributors, and the negative charge is evenly distributed between the two oxygen atoms.
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
The stabilization of the conjugate base pulls the position of equilibrium to the right, so the compound is more acidic than 1.
Answer:
The final volume will be 5.80 L
Explanation:
Step 1: Data given
Number of moles gas = 0.140 moles
Volume of gas = 2.78 L
Number of moles added = 0.152 moles
Step 2: Calculate the final volume
V1/n1 = V2/n2
⇒ with V1 = the initial volume = 2.78 L
⇒ with n1 = the initial number of moles = 0.140 moles
⇒ with V2 = The new volume = TO BE DETERMINED
⇒ with n2 = the new number of moles = 0.140 + 0.152 = 0.292 moles
2.78/0.140 = V2 /0.292
V2 = 5.80 L
The final volume will be 5.80 L