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den301095 [7]
2 years ago
13

What is the empirical formula for ribose (C5H10O5)?

Chemistry
1 answer:
Dafna1 [17]2 years ago
3 0

What is the empirical formula for ribose (C5H10O5)?

C. CH20

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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
6. Which part leads the blood to the heart?
Dmitriy789 [7]

Answer:

Your answer should be D

7 0
2 years ago
Read 2 more answers
Calculate the number of pounds of CO2CO2 released into the atmosphere when a 22.0 gallon22.0 gallon tank of gasoline is burned i
Gnesinka [82]

Answer:

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasoline be M

Density= \frac{Mass}{Volume}

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Given that gasoline is primarily octane.

2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O

Mass of octane burnt in the tank = M = 57,629.081 g

Moles of octane =\frac{57,629.081 g}{114.08g/mol}=505.1637 mol

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

\frac{16}{2}\times 505.1637 mol=4,041.3100 mol of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

3 0
3 years ago
How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.48 L against a constant pressure
Dahasolnce [82]
Given:
ΔV = 5 - 1.48 L = 3.52 L = 0.00352 m³, th chane in volume
p = 0.818 atm = 0.819*101325 Pa = 82985.2 Pa

By definition, the work done is
W = \int pdV = (82985.2 \,  \frac{N}{m^{2}} )*(0.00352 \, m^{3}) = 292.1 \, J

Answer: 292.1 J
5 0
3 years ago
The process by which water vapor changes to liquid water is called
andre [41]
When water is turned into vapor (a gas), that is called evaporation.

Hope this helps. :)
6 0
3 years ago
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