Draw 1,2,3,4,5,6-hexachlorocyclohexane with :a. all the chloro groups in axial positions.
b. all the chloro groups in equatorial positions.

Answers

Answer 1
Answer:

Answer:

This is required answer.

Explanation:

Given that,

1,2,3,4,5,6-hexachlorocyclohexane

(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.

When we say that all the chloro groups in axial position that means axial bonds are vertical.

(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions

Using given data

We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in  equatorial positions.

When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.

Hence, This is required answer.


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Be sure to answer all parts. Consider both 5-methyl-1,3-cyclopentadiene (A) and 7-methyl-1,3,5-cycloheptatriene (B). Which labeled H atom is most acidic? Hb is most acidic because its conjugate base is aromatic. Hc is most acidic because its conjugate base is antiaromatic. Ha is most acidic because its conjugate base is antiaromatic. Hd is most acidic because its conjugate base is aromatic. Which labeled H atom is least acidic? Ha is least acidic because its conjugate base is aromatic. Hb is least acidic because its conjugate base is antiaromatic. Hd is least acidic because its conjugate base is aromatic. Hc is least acidic because its conjugate base is antiaromatic.

Answers

Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

What are the most and the least acidic hydrogen atom?

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

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I don’t have a picture but I can describe it to you.

The hydrogen that is attached at the tertiary position on the heptatriene (at the 7-methyl) would be very acidic, as removal would leave a positive charge that could be moved throughout the ring through resonance. This would mean that the three double bonds would be participating in resonance, and the deprotonated structure would be aromatic, thus making this favorable.

The hydrogen that is attached at the tertiary position on the pentadiene (5-methyl) would NOT be acidic, as removal would cause an antiaromatic structure.

Any other hydrogens would NOT be acidic. Those vinylic to their respective double bonds would seriously destabilize the double bond if removed, and hydrogens attached to the methyl group jutting off the ring have no incentive to leave the carbon.

Hope this helps!

What is the difference between a volume that is delivered and a volume that is contained?

Answers

Answer:

When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.

A solution contains 1.694 mg CoSO4 (155.0 g/mol) per milliliter. Calculate (a) the volume of 0.08640 M EDTA needed to titrate a 25.00-mL aliquot of this solution. (b) the volume of 0.009450 M Zn2 needed to titrate the excess reagent after addition of 50.00 mL of 0.08640 M EDTA to a 25.00-mL aliquot of this solution.

Answers

Moles of any substance is given by the molar mass and the mass. The volume of EDTA required to titrate is 3.16 mL and the volume of zinc required to titrate is 22.8 mL.

What is molarity?

Molarity is the property of the solution that gives the concentration of the solute present in the solution.

Given,

Mass of cobalt sulfate = 1.697 gm

The molar mass of cobalt sulfate = 155 g/mol

In the first part, the volume of the aliquot is 25 mL and the molarity is 0.08640 M.

The reaction is shown as:

Co²⁺ + H₄Y → CoH₂Y + 2H⁺

Moles of cobalt: n = 0.001694 ÷ 155 = 0.0000109 moles

In 25 ml aliquot moles of cobalt are, 0.000273 moles.

The volume of EDTA solution is calculated as:

V = moles ÷ Molarity

= 0.000273 mole ÷ 0.0864 mol/ L

= 3.16 mL

Hence, 3.16 mL of EDTA is required.

For the second part, moles of EDTA are calculated as:

n = 0.008640 × 0.050 = 4.32 ×10⁻⁴

In a 25 mL, sample moles of EDTA are 2.16 × 10⁻⁴ moles.

Excess moles of EDTA in the solution:

0.000432 - 0.000216 = 0.000216 moles

The volume of EDTA is calculated as:

V = 0.000216 ÷ 0.009450

= 0.0228 L

Hence, 22.8 mL of zinc is required.

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Answer:

the answer is in the screenshot

Explanation:

hope this helps

A compound contains only carbon, hydrogen, and oxygen. Combustion of 139.1 g of the compound yields 208.6 g of CO2 and 56.93 g of H2O. The molar mass of the compound is 176.1 g/mol. *Each part of this problem should be submitted separately to avoid losing your work* 1. Calculate the grams of carbon (C) in 139.1 g of the compound: grams 2. Calculate the grams of hydrogen (H) in 139.1 g of the compound. grams 3. Calculate the grams of oxygen (O) in 139.1 g of the compound. grams

Answers

Answer:

1. Mass of Carbon is 56.89g

2. Mass of Hydrogen is 6.33g

3. Mass of Oxygen is 75.88

Explanation:

The following were obtained from the question.

Mass of the compound = 139.1g

Mass of CO2 produced = 208.6g

Mass of H2O produced = 56.93

1. Determination of mass of Carbon (C). This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 208.6

Mass of C = 56.89g

2. Determination of the mass of Hydrogen (H). This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 56.93

Mass of H = 6.33g

3. Determination of the mass of oxygen (O).

This is illustrated below:

Mass of the compound = 139.1g

Mass of C = 56.89g

Mass of H = 6.33g

Mass of O = Mass of compound - (mass of C + Mass of H)

Mass of O = 139.1 - (56.89 + 6.33)

Mass of O = 139.1 - 63.22

Mass of O = 75.88

Which compound contains covalent bonds?A - BeO
B- NaCl
C- CaS
D- MgF2
E - SO2

Answers

Answer:

A

Explanation:

Thus is because both the elements are non metals.

Answer:

SO2

Explanation:

SO2 is covalent since it is formed by the sharing of electrons between sulphur and oxygen also both the sulphur and oxegen atoms are non metal.

and we know that non-metal non-metal always bond together via covelent bonding

What is the empirical formula of a compound composed of 36.7 g 36.7 g of potassium ( K K ) and 7.51 g 7.51 g of oxygen ( O O )? Insert subscripts as needed.

Answers

Answer:

K₂O

Explanation:

Given parameters:

Mass of K = 36.7g

Mass of O = 7.51g

Unknown:

Empirical formula of the compound

Solution:

The empirical formula of a compound is it's simplest ratio by which the elements in the compound combines. It differs from the molecular formula that shows the actual atomic ratios.

To find the empirical formula, follow this process;

   

Elements                     K                               O

Mass                          36.7                            7.51

Molar

mass                            39                            16

Number of

moles                       36.7/39                    7.51/16

                                   0.94                        0.47

Divide by

the smallest              0.94/0.47               0.47/0.47

                                         2                              1

  Empirical formula is        K₂O

                                     

Final answer:

The empirical formula of the compound composed of 36.7 g of potassium and 7.51 g of oxygen is K2O.

Explanation:

To determine the empirical formula of a compound, we need to find the ratio of the elements present. In this case, we have 36.7 g of potassium and 7.51 g of oxygen. To find the ratio, we need to convert these masses to moles by dividing them by the molar masses of potassium and oxygen. The molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Dividing the masses by the molar masses gives us 0.939 mol potassium and 0.469 mol oxygen. The ratio between these two elements is approximately 2:1, so the empirical formula of the compound is K2O.

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