Answer:
Option 1, Cl is reduced and gains electrons
Explanation:
HClO₃ → HClO₂
In HClO₃, chlorine acts with +5 in the oxidation state
In HClO₂,, chlorine acts with +3 in the oxidation state.
The state has been reducted, so the Cl has been reduced. As it was reduced, it means that has won e⁻, in this case 2
Cl⁻⁵ → Cl⁻³ + 2e⁻
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
Answer: Potassium hydroxide, KOH, is considered a <u>BASE</u> in an acid-base reaction because it <u>ACCEPTS</u> a hydrogen ion from the other reactant.
According to Brønsted–Lowry acid–base theory, Base is a specie which accepts proton (H⁺) while, Acid is a specie which donate proton.
Bases may contain a negative charge or lone pair of electrons, while, Acids contain positive charge or a neutral atom with incomplete octet.
In given statement KOH is acting as a base because it contains a negatively charged hydroxyl group which can accept proton from a acid, i.e.
KOH → K⁺ + OH⁻
Reaction of OH⁻ with any acid,
K⁺ + OH⁻ + HCl → H₂O + KCl
Answer : The equilibrium concentration of 
at 
is, 
.
Solution : Given,
Equilibrium constant, 
Initial concentration of 
= 0.260 m
Let, the 'x' mol/L of are formed and at same time 'x' mol/L of are also formed.
The equilibrium reaction is,
Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
![K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get
By rearranging the terms, we get the value of 'x'.
Therefore, the equilibrium concentration of at is, .
