The fumes of acetic acid go to the floor instead of ceiling.
The reason behind the answer is that the density of acetic acid fumes is higher than the density of air (almost twice) that is equal to only 1
.
Thus, the fumes of acetic acid go to the floor.
Answer:
Nandiyan yan sa binasa mo
Answer:
A. NaHCO₃
Explanation:
NaHCO₃ ⇒ NaOH + H₂CO₃
NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.
Hope that helps.
Answer:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Explanation:
A → B + C
The rate law of the reaction will be :
![R=k[A]^x](https://tex.z-dn.net/?f=R%3Dk%5BA%5D%5Ex)
Initial rate of the reaction when concentration of the substrate was 0.4 M:
..[1]
Initial rate of the reaction when concentration of the substrate was 0.8 M:
...[2]
[1] ÷ [2] :
![\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}](https://tex.z-dn.net/?f=%5Cfrac%7B0.183%20M%2Fs%7D%7B0.183%20M%2Fs%7D%3D%5Cfrac%7Bk%5B0.4%20M%5D%5Ex%7D%7Bk%5B0.8%20M%5D%5Ex%7D)
x = 0
The order of the reaction is zero.
For the value of rate constant ,k:
..[1]
x = 0
![0.183 M/s=k[0.4 M]^0](https://tex.z-dn.net/?f=0.183%20M%2Fs%3Dk%5B0.4%20M%5D%5E0)
k= 0.183 M/s
The half life of the zero order kinetics is given by :
![t_{1/2}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)
Where:
= Initial concentration of A
k = Rate constant of the reaction
So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:
![t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B2.77%20M%7D%7B0.183%20M%2Fs%7D%3D15.1%20s)
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.