Answer : (b) The rate law expression for the reaction is:
![\text{Rate}=k[SO_2]^2[O_2]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BSO_2%5D%5E2%5BO_2%5D)
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
The general reaction is:
The general rate law expression for the reaction is:
![\text{Rate}=k[A]^a[B]^b](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BA%5D%5Ea%5BB%5D%5Eb)
where,
a = order with respect to A
b = order with respect to B
R = rate law
k = rate constant
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
and ![[B]](https://tex.z-dn.net/?f=%5BB%5D)
= concentration of A and B reactant
Now we have to determine the rate law for the given reaction.
The balanced equations will be:
In this reaction, 
and 
are the reactants.
The rate law expression for the reaction is:
![\text{Rate}=k[SO_2]^2[O_2]^1](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BSO_2%5D%5E2%5BO_2%5D%5E1)
or,
Answer:
pH = 12.08
Explanation:
First we <u>calculate how many moles of each substance were added</u>, using <em>the given volume and concentration</em>:
- HBr ⇒ 0.05 M * 75 mL = 3.75 mmol HBr
- KOH ⇒ 0.075 M * 74 mL = 5.55 mmol KOH
As HBr is a strong acid, it dissociates completely into H⁺ and Br⁻ species. Conversely, KOH dissociates completely into OH⁻ and K⁺ species.
As there are more OH⁻ moles than H⁺ moles (5.55 vs 3.75), we <u>calculate how many OH⁻ moles remain after the reaction</u>:
- 5.55 - 3.75 = 1.8 mmoles OH⁻
With that<em> number of moles and the volume of the mixture</em>, we <u>calculate [OH⁻]</u>:
- [OH⁻] = 1.8 mmol / (75 mL + 74 mL) = 0.0121 M
With [OH⁻], we <u>calculate the pOH</u>:
With the pOH, we <u>calculate the pH</u>:
A: ionic salt
i've done some research and i believe it's A
Answer:
aaaaahhhhhhhhhhhhhhhhhhhhhhhhhh
Explanation:
Answer:a line or a circle but i think circle
Explanation:
