Answer: a) The concentration after 8.8min is 0.17 M
b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.
Explanation:
Expression for rate law for first order kinetics is given by:

where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) concentration after 8.8 min:



b) for concentration to decrease from 0.25M to 0.15M


Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
Hello!
To solve this problem, we will use the
Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

So, the final volume after increasing the pressure would be
2,7 L. That means that volume decreases when the pressure increases
Have a nice day!
Human:
✧・゚: *✧・゚:* Answer: *:・゚✧*:・゚✧
✅ the first one
I WOULD APPRECIATE BRAINLIEST!
~ ₕₒₚₑ ₜₕᵢₛ ₕₑₗₚₛ! :₎ ♡
~
I believe it is B. I’m sorry if i’m wrong. Tell me if i’m wrong or right!