Answer:
[HI] = 0.097 M
Explanation:
Let's consider the following reaction.
2 HI(g) ⇄ H₂(g) + I₂(g)
The order of reaction for HI is 1. Thus, we can calculate the concentration of HI ([HI]) at certain time using the following expression:
ln [HI] = ln [HI]₀ - k. t
where,
[HI]₀ is the initial concentration of HI
k is the rate constant
t is the time elapsed
When [HI]₀ = 0.440 M and t = 0.210 s, the concentration of HI is
ln [HI] = ln (0.440) - 7.21 s⁻¹ × 0.210 s
ln [HI] = -2.33
[HI] = 0.097 M
Answer:
yes
Explanation:
The molecular formula for a compound can be the same as or a multiple of the compound's empirical formula. Molecular formulas are compact and easy to communicate; however, they lack the information about bonding and atomic arrangement that is provided in a structural formula.
Answer:
make contact with the heat source
Explanation:
Answer:
Explanation:
To find how many moles are in the sample, you first must calculate how much one mole of C2H5OH weighs.
2(mass of C) + 5(mass of H) + (mass of O) + (mass of H)
= 2(12.01) + 5(1.008) + (16.00) + (1.008) = 46.068 g/mol
So, now that we know what 1 mole weighs, we can simply divide the weight of the sample by the weight of a mole of ethanol to get our answer.
39.2 / 46.068 = approximately 0.851 mol of ethanol
Answer:
1 = oxidation
2 = reduction
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
2I- ----> I₂+ 2e⁻
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
F + e⁻ ----> F⁻
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.