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Fantom [35]
3 years ago
11

A 30 wt % solution of NaOH is diluted in a mixer to 5 wt%. If the streams entering and leaving the mixer are at 40 oC, find the

heat removed from the mixer on a basis of 100 kg of feed solution.
Chemistry
1 answer:
Strike441 [17]3 years ago
6 0

Explanation:

It is given that 100 kg feed solution  contains 30 wt% solution NaOH.  Hence, feed contains 70 kg of water and 30 kg of NaOH.

Therefore, mixer outlet shows 5 wt% NaOH.

Let us assume that the mixture outlet be F kg  so, by applying mass balance of NaOH we get the value of force as follows.

          F \times 0.05 = 30

                 F = 600 Kg

In that 30 kg is NaOH and 570 kg is water  which also means that initially water present is 70 kg. And, additional water added is 500 kg .

Thus, water feed rate is 500 kg/hr.

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Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution. H+ + H2O2 ? H3
Margarita [4]

<u>Answer:</u> The rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

<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:

2H^++2Br^-+H_2O_2\rightarrow Br_2+2H_2O

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  H^++H_2O_2\rightleftharpoons H_3O_2^+;\text{ (fast)}

<u>Step 2:</u>  H_3O_2^++Br^-\rightarrow HOBr+H_2O;\text{(slow)}

<u>Step 3:</u>  HOBr+H^++Br^-\rightarrow Br_2+H_2O;\text{(fast)}

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^-]          ......(1)

As, [H_3O_2^+] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for [H_3O_2^+] from step 1, we get:

K=\frac{[H_3O_2^+]}{[H^+][H_2O_2]}  

[H_3O_2^+]=K[H^+][H_2O_2]

Putting the value of [H_3O_2^+] in equation 1, we get:

\text{Rate}=k.K[H^+][H_2O_2][Br^-]\\\\\text{Rate}=k'[H+][H_2O_2][Br^-]

Hence, the rate law for the reaction is \text{Rate}=k'[H+][H_2O_2][Br^-]

4 0
3 years ago
A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the bl
lbvjy [14]

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

Download odt
3 0
3 years ago
Why it is advisided to freshly prepare the ferrous sulphate solution for experiment
chubhunter [2.5K]
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
5 0
3 years ago
In each reaction box place the best reagent and conditions from the list below benzene 3 boxes
den301095 [7]

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.

Benzene     \frac{AlCl3}{Ch3Cl}>   Toulene + HCl

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 Tolune   \frac{HNO3 -H2SO4}{30-40 degree C} ->  2,4- dinitrotoluene

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 2,4 -dinitrotoluene   \frac{fuming HNO3-H2So4}{90-100 C} ->  2,4,6-trinitrotoluene.

So over all reaction uses three reagents in order :

Benzene  \frac{AlCl3}{CH3Cl}  -> Toluene  \frac{HNO3-H2So4}{room temp}  -> 2,4-dinitrotoluene  \frac{Fuming HNO3 -H2SO4}{Heating at 90-100 C}  -> 2,4,6-trinitrotoluene .

3 0
3 years ago
Read 2 more answers
Different substances have....
Sergio039 [100]

Answer:

Different properties

Explanation:

They are different.

8 0
2 years ago
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