Out of the options, "Iron" is the most dense here.
In short, Your Answer would be Option D
Hope this helps!
Answer:
uwu
Explanation:
Because I like UwU cats, are you one?
Answer:
The answer to your question is below
Explanation:
1)
Balanced chemical reaction
2CH₃OH + 3O₂ ⇒ 2 CO₂ + 4H₂O
Reactant Element Product
2 C 2
8 H 8
8 O 8
Molar mass of CH₃OH = 2[12 + 16 + 4]
= 2[32]
= 64 g
Molar mass of O₂ = 3[16 x 2] = 96 g
Theoretical proportion CH₃OH/O₂ = 64 g/96g = 0.67
Experimental proportion CH₃OH/O₂ = 60/48 = 1.25
Conclusion
The limiting reactant is O₂ because the Experimental proportion was higher than the theoretical proportion
2)
Balanced chemical reaction
S₈ + 12O₂ ⇒ 8SO₃
Reactant Elements Products
8 S 8
24 O 24
Molar mass of S₈ = 32 x 8 = 256 g
Molar mass of O₂ = 12 x 32 = 384 g
Theoretical proportion S₈ / O₂ = 256 / 384
= 0.67
Experimental proportion S₈ / O₂ = 40 / 35
= 1.14
Conclusion
The limiting reactant is O₂ because the experimental proportion was lower than the theoretical proportion.
Answer:
A) 6.48 g of OF₂ at the anode.
Explanation:
The gas OF₂ can be obtained through the oxidation of F⁻ (inverse reaction of the reduction presented). The standard potential of the oxidation is the opposite of the standard potential of the reduction.
H₂O(l) + 2 F⁻(aq) → OF₂(g) + 2 H⁺(aq) + 4 e⁻ E° = -2.15 V
Oxidation takes place in the anode.
We can establish the following relations:
- 1 Faraday is the charge corresponding to 1 mole of e⁻.
- 1 mole of OF₂ is produced when 4 moles of e⁻ circulate.
- The molar mass of OF₂ is 54.0 g/mol.
The mass of OF₂ produced when 0.480 F pass through an aqueous KF solution is:
