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2 years ago

A hydrocarbon substituent is called an

Chemistry

2 answers:

saw5 [17]2 years ago

5 0

It’s called An alkyl group.

vekshin12 years ago

4 0

Aliphatic hydrocarbons are divided into three main groups according to the types of bonds they contain: alkanes, alkenes, and alkynes. Alkanes have only single bonds, alkenes contain a carbon-carbon double bond, and alkynes contain a carbon-carbon triple bond.

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4. Synthesis problems. Propose a synthetic route from the starting material to the product. In your answer, show each individual

Troyanec [42]

The reduction of alkyne to an alkene in the first step allows the best reagent to be chosen for each subsequent step.

Describe reagents.

A reagent is merely an essential component of a chemical reaction, it should be mentioned. It is an ingredient that speeds up the reaction.

With H2 and Lindlar's catalyst, an alkyne is reduced to alkene as the initial step in this process. Alkene will then be brominated to produce allyl bromide as the next step.

In this instance, the required allyl alcohol will be produced via the reaction of allyl bromide with NaOH.

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1 year ago

Glycerol boils at a high temperature than water. What does this indicate about the attractive forces of glycerol?

forsale [732]

Glycerol attractive forces are great than water. The harder to break, the more energy is needed.

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3 years ago

When 50,0 g of silicon dioxide is heated with an excess of carbon, 32.2 g of

DerKrebs [107]

This reaction is called silioconic

8 0

3 years ago

Match the following associations. 1. + ΔHexothermic endothermic 2. E2 < E¹exothermic endothermic 3. energy is absorbed exothe

Alisiya [41]

Explanation:

When heat is absorbed by reactants in a chemical reaction then the reaction is known as endothermic reaction.

Whereas when heat is released by reactants in a chemical reaction then the reaction is known as exothermic reaction.

1. When $\Delta H$ is positive then it means heat is absorbed by reactants and reaction is endothermic.

2. When $E_{2} < E_{1}$ , this means energy of products is less than the energy of reactants. Therefore, it is exothermic in nature.

3. When energy is absorbed then reaction is endothermic.

4. When $E_{2} > E_{1}$ , this means energy of products is greater than the energy of reactants. Therefore, it is endothermic in nature.

5. When [tex]\Delta H is negative then it means heat is released by reactants and reaction is exothermic.

4 0

3 years ago

Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a

kolezko [41]

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) = $25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)$

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= $25 g x \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}$

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x $\frac{40.67 KJ}{mol H20}$ x $\frac{1 mol H20}{18 g}$

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

3 0

3 years ago