<u>Answer:</u> The rate law for the reaction is ![\text{Rate}=k'[H+][H_2O_2][Br^-]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%27%5BH%2B%5D%5BH_2O_2%5D%5BBr%5E-%5D)
<u>Explanation:</u>
Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.
In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.
The chemical equation for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution follows:
The intermediate reaction of the mechanism follows:
<u>Step 1:</u> 
<u>Step 2:</u> 
<u>Step 3:</u> 
As, step 2 is the slow step. It is the rate determining step
Rate law for the reaction follows:
![\text{Rate}=k[H_3O_2^+][Br^-]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BH_3O_2%5E%2B%5D%5BBr%5E-%5D)
......(1)
As, ![[H_3O_2^+]](https://tex.z-dn.net/?f=%5BH_3O_2%5E%2B%5D)
is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.
Applying steady state approximation for from step 1, we get:
![K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_3O_2%5E%2B%5D%7D%7B%5BH%5E%2B%5D%5BH_2O_2%5D%7D)
![[H_3O_2^+]=K[H^+][H_2O_2]](https://tex.z-dn.net/?f=%5BH_3O_2%5E%2B%5D%3DK%5BH%5E%2B%5D%5BH_2O_2%5D)
Putting the value of in equation 1, we get:
![\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk.K%5BH%5E%2B%5D%5BH_2O_2%5D%5BBr%5E-%5D%5C%5C%5C%5C%5Ctext%7BRate%7D%3Dk%27%5BH%2B%5D%5BH_2O_2%5D%5BBr%5E-%5D)
Hence, the rate law for the reaction is
Answer:
For iron
Final temperature = 54,22°C
For copper
Final Temperature = 63.67 °C
Explanation
Hello,
You are using a torch to warm up a block of iron that has an initial temperature of 32°C.
The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.
So you need to use the heat capacity equation as follow in the paper.
The equation has to have all terms in the same units, so:
q = 12000 J
s = 0.450 J / g °C
m = 1200 g
Ti = 32 °C
Here is a site my buddie has to help you. Well co-owner..
https://www.quora.com/Why-is-fresly-prepared-FeSO4-required-for-the-ring-test
Answer : The correct answer is 1) AlCl₃ - CH₃Cl 2) HNO₃ -H₂SO₄ at room temperature 3) Fuming HNO₃ -H₂SO₄ at 90-100 ⁰ C heat .
I think this reaction is forming 2,4,6- trinitrotoluene from benzene, since the product is not mentioned. Following are the steps to convert Benzene to 2,4,6 trinitrotoluene .
Step 1: Conversion of Benzene to Toluene .
Benzene can be converted to toluene by Friedel Craft Alkylation of benzene . In this reaction reagent AlCl₃ and Ch3Cl is used . Electrophile CH³⁺ is produced which attached on carbon of benzene and formation of Toluene and HCl occur.
Step 2 : Conversion of Toluene to dinitrotoluene.
Dinitritoluene is prepared from toluene by Nitration . This reaction uses Electrophilic substitution mechanism . The reagents used are HNO₃ and H₂SO₄ at room temperature . These reagents produces NO₂⁺ ( nitronium ion ), a electrophile which attacks on C2 and C4 Carbon atoms of Toluene.
Toluene 
Step 3) Conversion of Dinitro toluene to trinitrotoluene.
This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.
Dinitrotoluene 
So over all reaction uses three reagents in order :
