Nonmetals and negative. Answer is c
Answer:
the answer is D i believe. i am not shure but im like 70% sure
Answer:
The concentration of cyclobutane after 875 seconds is approximately 0.000961 M
Explanation:
The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M
The final concentration of cyclobutane, [
] = 0.00119 M
The time for the reaction, t = 455 seconds
Therefore, the Rate Law for the first order reaction is presented as follows;
![\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }](https://tex.z-dn.net/?f=%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D%20%5Ctext%20%7B-k%7D%20%5Ccdot%20t%20%7D)
Therefore, we get;
![k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%7D%20%20%7B-t%20%7D)
Which gives;

k ≈ 5.088 × 10⁻⁴ s⁻¹
The concentration after 875 seconds is given as follows;
[
] = [A₀]·
Therefore;
[
] = 0.00150 ×
= 0.000961
The concentration of cyclobutane after 875 seconds, [
] ≈ 0.000961 M
Answer:
0.355 N of HF
Explanation:
The titration reaction of HF with KOH is:
HF + KOH → H₂O + KF
<em>Where 1 mole of HF reacts per mole of KOH</em>
<em />
Moles of KOH are:
0.0296L × (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.
As volume of the titrated solution was 10.0mL, normality of HF solution is:
3.552x10⁻³ equivalents of HF / 0.010L =<em> 0.355 N of HF</em>