Balanced chemical reaction: 2HI(aq) + Ba(OH)₂(aq) → BaI₂(aq) + 2H₂O(l).
Ionic reaction: 2H⁺ + I⁻(aq) + Ba²⁺ + 2OH⁻(aq) → Ba²⁺ + 2I⁻(aq) + 2H₂O(l).
Net ionic reaction: 2H⁺ + 2OH⁻(aq) → 2H₂O(l).
Barium iodide is salt that dissolves in water, barium hydroxide is strong base that dissolves in water.
This is example of double replacement reactions(double displacement or metathesis reactions), two ionic compounds are exchanged, making two new compounds
.
Answer:
The expected ratio of half-lives for a reaction will be 5:1.
Explanation:
Integrated rate law for zero order kinetics is given as:
![k=\frac{1}{t}([A_o]-[A])](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%28%5BA_o%5D-%5BA%5D%29)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial concentration
[A]=concentration at time t
k = rate constant
if, ![[A]=\frac{1}{2}[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B1%7D%7B2%7D%5BA_o%5D)
, the equation (1) becomes:
![t_{\frac{1}{2}}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)
Half life when concentration was 0.05 M=
Half life when concentration was 0.01 M=
Ratio of half-lives will be:
![\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}](https://tex.z-dn.net/?f=%5Cfrac%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%27%7D%3D%5Cfrac%7B%5Cfrac%7B%5B0.05%20M%5D%7D%7B2k%7D%7D%7B%5Cfrac%7B%5B0.01%20M%5D%7D%7B2k%7D%7D%3D%5Cfrac%7B5%7D%7B1%7D)
The expected ratio of half-lives for a reaction will be 5:1.
<span>Air enters through the nose or mouth (which join to form the nasopharyngeal cavity). The air travels through the trachea which and the trachea splits into two bronchi. The air travels through the bronchi which split into smaller and smaller bronchioles. The tiny bronchioles and the air ends up in the miniscule alveoli, where the oxygen in the air diffuse into the bloodstream. Carbon dioxide diffuse from the bloodstream into the alveoli and the unwanted gas travels in reverse back to the nose and mouth, where it is breathed out.</span>
Answer is: <span>the molarity of the diluted solution 0,043 M.
</span>V(NaOH) = 75 mL ÷ 1000 mL/L = 0,075 L.
c(NaOH) = 0,315 M = 0,315 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,075 L · 0,315 mol/L.
n(NaOH) = 0,023625 mol.
V(solution) = 0,475 L + 0,75 L.
c(solution) = 0,023625 mol ÷ 0,550 L.
c(solution) = 0,043 mol/L.
