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mafiozo [28]
3 years ago
15

In each reaction box place the best reagent and conditions from the list below benzene 3 boxes

Chemistry
2 answers:
blondinia [14]3 years ago
8 0

The best reagent and conditions from the list below benzene 3 boxes are as attached in the picture below

<h3>Explanation: </h3>

In each reaction box place the best reagent and conditions from the list below benzene 3 boxes.

A reagent is a substance or compound added to a system causing a chemical reaction. Whereas Benzene is an organic chemical compound with the chemical formula C6H6.

Br2 (Diatomic Bromine) , FeBrg AICI: CI . AICI Br2 CH3CH2CH2CI, AICI HNO3, H2SO4 NBS, ROOR, hv NaBH4 AICI H2 Zn(Hg), HCI CH3CH2CI, AICI

  • Aluminum chloride is a chemical compound with the chemical formula AlCl3. It is mainly produced and consumed in the production of aluminium metal, but large amounts are also used in other areas of the chemical industry.
  • BM2 presents a unique opportunity to understand how different chemical systems perform similar functions.
  • Zinc amalgam (Zn-Hg) has one important use: in the Clemmensen reduction of ketones to alkanes. The definition of an amalgam is a mixture of metals and mercury, that can be man-made or can be naturally occurring. The example of an amalgam is a mixture of silver and mercury which is used as a dental filling.
  • Hydrochloric acid is the aqueous solution of hydrogen chloride. It is a strong mineral acid with many industrial uses.

Learn more about Aluminum chloride brainly.com/question/12053181

#LearnWithBrainly

den301095 [7]3 years ago
3 0

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 .

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Answer:

mCO2= 49.6932 kgCO2

Explanation:

Hello! Let's solve this!

First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O

We see that each mole of C3H8 (propane) we get 3 moles of CO2

From the propane volume we can obtain the grams of propane used.

molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane

mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2

mCO2= 49.6932 kgCO2

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Determine which of these properties would distinguish these two substances: (a) boiling point; (b) combustion analysis results;
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Answer:

(a) boiling point

(d) density at a given temperature and pressure.

Explanation:

Isomers are compounds that have the same molecular formula but different structural formulas. They differ in chemical and physical properties depending on the type of isomerism displayed by the compounds.

The compounds stated here are structural or constitutional isomers hence they possess different boiling points and densities at a given temperature and pressure owing to structural differences in the molecules.

Since they have the same molecular formula, they must yield the same result during combustion analysis and they must have the same molecular weight.

3 0
3 years ago
The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.
Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0
3 years ago
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