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topjm [15]
3 years ago
6

Draw the structural formula of (3e,5e)-2,5-dibromo-3,5-octadiene.

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
4 0

We need draw the structure of (3E,5E)-2,5-dibromo-3,5-octadiene.

The structure is shown below:

The molecule has two double bonds in 3 and 5 position and two Br-atoms attached to 2 and 5 position.

The configuration is 3E and 5E. E configuration indicates cis- isomers where similar groups are in same side with respect to double bonds.

You might be interested in
What is an atom with 2 protons and 4 neutrons
Katena32 [7]
Helium,it has an atomic mass of 4,which means total no. of protons and neutrons,so I think you meant 2 protons and an atomic mass of 4
7 0
3 years ago
What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2
atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0
3 years ago
Read 2 more answers
A shiny chunk of metal is found to have a mass of 37.28g. The metal is dropped into a graduated cylinder which contains 20.0 mL
Amanda [17]

Answer: The density of the material is 2.66 g/mL and it is likely this  is made of Aluminum

Explanation:

The first step to know the material of the chunk of metal is to calculate its density. The general formula for density is P (density) = \frac{m (mass)}{ v (volume)}. Moreover, in this case, it is known the mass is 37.28 g, but the volume is not directly provided. However, we know the water in the graduated cylinder had a volume of 20.0 mL and this increased to 34.0 mL when the chunk of metal is added, this means the volume of the metal is 14 mL (34.0 mL - 20.0 mL = 14 mL). Now let's calculate the density:

P = \frac{37.28g}{14.0mL}

P = 2.66 g/mL

This means the density of this metal is 2.66 g/mL, which can be rounded as 2. 7 g/mL, and according to the chart, this is the density of aluminum. Therefore, this material of this chunk is aluminum.

6 0
4 years ago
1. How will you differentiate physical from chemical change in terms of the
Tpy6a [65]

Answer:

In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed.

3 0
3 years ago
12. A helium-filled weather balloon has a volume of 2.4 x 102 L at 99 kPa pressure and a
Luba_88 [7]

Answer:

The answer to your question is 8.74 g of He

Explanation:

Data

V = 2.4 x 10² L

P = 99 kPa

T = 0°C

mass = ?

Process

1.- Convert kPa to atm

P = 99 kPa = 99000 Pa

                   1 atm --------------- 101325 Pa

                    x       ---------------   99000 Pa

                   x = (99000 x 1) / 101325

                   x = 0.977 atm

2.- Convert temperature to °K

°K = 273 + 0

°K = 273

3.- Substitution

      PV = nRT

- Solve for n

      n = PV / RT

      n = (0.977)(2.4 x 10²) / (0.082)(273)

      n = 24.48 / 22.386

      n = 1.093 moles

4.- Calculate the grams of He

          8 g -------------------- 1 mol

           x    -------------------- 1.093 moles

           x = (1.093 x 8) / 1

           x = 8.74 g                      

6 0
4 years ago
Read 2 more answers
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