The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxidation of trans-2-butene. Draw the products and explain why the configuration of the starting alkene is not relevant in this case.

Answers

Answer 1

Answer:

a) Attached below

b)  The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration

Explanation:

Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.

attached below

The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration )

The Products Obtained From Hydroboration-oxidation Of Cis-2-butene Are Identical To The Products Obtained

Related Questions

Will give brainliest answer please give explanation


33.5 cs=_________s

Answers

Answer: 3.4 × 10 -1 equal 33.5 cs. May I have brainiest? pls. (the -1 is to the power of the ten.

Give the symbol for an element that is:__________

a. a halogen: _______________
b. an alkali metal: _______________
c. a noble gas: _______________
d. an alkaline earth metal : ____________

Answers

a. Br, Cl, F

b. Na, K, Ba

c. He, Ar, Ne

d. Ca, Ba, Mg

Answer:

a. halogen : F ,Cl ,Br l ,At

b an alkali metal: Na,Li, Rb, Cs

c. a noble gas: He, Ne, Kr, Ar

d. an alkaline earth metal: Be,Mg,Ca, Sr

hope it helps

stay safe healthy and happy...

Explain why caffeine can be extracted from the tea leaves into hot water and how you extracted the aqueous solution and isolated the crude caffeine.

Answers

Answer:

The hot water dissolves the flavor and color components.

Explanation:

Caffeine can be extracted from the tea leaves into hot water because the hot water dissolves the flavor and color components away from the solid vegetable. This is an example of a solid-liquid extraction. We can extracted the aqueous solution and isolated the crude caffeine by converting the components of caffeine into their calcium salts which are insoluble in water. Then the caffeine can be extracted from the water by using methylene chloride.

A sample of Br2(g) takes 12.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane

Answers

Answer:

6 mins

Explanation:

The time taken for Ar to effuse can be obtained as follow:

Time for Br₂ (t₁) = 12 mins

Molar mass of Br₂ (M₁) = 2 × 80 = 160 g/mol

Molar mass of Ar (M₂) = 40 g/mol

Time for Ar (t₂) =?

t₂/t₁= √(M₂/M₁)

t₂ / 12 = √(40/160)

Cross multiply

t₂ = 12 × √(40/160)

t₂ = 12 × 0.5

t₂ = 6 mins

Therefore, it will take 6 mins for the same amount of Ar to effused out.

What is the law of multiple proportions?
A. The proportion of elements to compounds is constant.
B. All elements are found in equal proportions in nature.
C. Different compounds may contain the same elements but may have different ratios of those elements.
D. All compounds contain the same elements in the same proportions.

Answers

Answer:

I think

(d) All compounds contain the same elements in the same properties

Consider the balanced chemical equation below.

3 A ⟶ C + 4 D

How many moles of D would be produced if 9 moles of A were used?

Answers

hope the picture above helps :)

Question 6: Energy Use (8 points)

A. The electricity supply of a certain U.S. state needs to be increased. The state is very cloudy and rainy from October to April. There are several large rivers, and there are almost no areas with high winds due to heavy forest cover. Several sources of underground heat are available. The state also has a surplus of organic waste.

i. Name two renewable energy sources, and state and explain whether or not the state should use them based on the description above. (6 points)

ii. If the state wants to minimize environmental damage, what energy sources should it consider using? Explain your position. (2 points)

Answers

Answer:

somebody small boy is not w e a r d

z0 0m = 257 473 5835 p c:- 12 34

Mava=mbvb ma x5.0ml = 5.2ml x 0.10m

Answers

Answer:

[tex]M_{a}[/tex] = 0.104 m

Explanation:

This expression is used to determine either the mass or volume of an acid or a base used during titration process.

So that;

[tex]M_{a}[/tex][tex]V_{a}[/tex] = [tex]M_{b}[/tex][tex]V_{b}[/tex]

[tex]M_{a}[/tex] is the mass of the acid

[tex]V_{a}[/tex] is the volume of the acid

[tex]M_{b}[/tex] is the mass of the base

[tex]V_{b}[/tex] is the volume of the base

Given that:

[tex]M_{a}[/tex] x 5.0 m l = 5.2 ml x 0.10 m

[tex]M_{a}[/tex] x 5.0 = 0.52 m

[tex]M_{a}[/tex] = [tex]\frac{0.52}{5.0}[/tex]

      = 0.104

[tex]M_{a}[/tex] = 0.104 m

The mass of the acid used is 0.104 m.

A gas bottle contains 0.650 mol of gas at 730. mmHg pressure. If the final pressure is 1.15 atm, how many moles of gas were added to the bottle

Answers

Answer: There are 0.779 moles of gas were added to the bottle.

Explanation:

Given: [tex]n_{1}[/tex] = 0.650 mol,     [tex]P_{1}[/tex] = 730 mm Hg (1 mm Hg = 0.00131579 atm) = 0.96 atm

[tex]n_{2}[/tex] = ?,           [tex]P_{2}[/tex] = 1.15 atm

Formula used is as follows.

[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}\\\frac{0.96 atm}{0.650 mol} = \frac{1.15 atm}{n_{2}}\\n_{2} = 0.779 mol[/tex]

Thus, we can conclude that there are 0.779 moles of gas were added to the bottle.

he equation for the dissociation of pyridine is
C5H5N(aq) + H2O(l) ⇌ C 5H5NH+(aq) + OH-(aq) Kb = 1.9 × 10-9

Calculate the pH of a pyridine solution that has a concentration of 9.2 M. Round your answer to two decimal places.

Answers

Answer:

10.10

Explanation:

Step 1: Write the basic dissociation reaction for pyridine

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq)      Kb = 1.9 × 10⁻⁹

Step 2: Calculate [OH⁻]

For a weak base, we will use the following expression.

[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M

Step 3: Calculate pOH

We will use the definition of pOH.

pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9

Step 4: Calculate pH

We will use the following expression.

pH = 14 - pOH = 14 - 3.9 = 10.10

Tartaric acid is the white, powdery substance that coats sour candies such as Sour Patch Kids. Combustion analysis of a 12.01-gg sample of tartaric acid, which contains only carbon, hydrogen, and oxygen, produced 14.08 gg CO2CO2 and 4.32 gg H2OH2O. Part A Find the empirical formula for tartaric acid. Express your answer as a chemical formula. Enter the elements in the order C, H, and

Answers

Answer:

C2H3O3

Explanation:

Empirical formula is the simplest whole number ratio of moles of atoms that you can find in a molecule.

In combustion analysis all Carbon reacts producing CO2 and all hydrogen reacts producing H2O. With the differences in masses we can find the mass of oxygen and their moles:

Moles CO2 = Moles C:

14.08g * (1mol/44.01g) = 0.3199 moles C * (12.01g/mol) = 3.8423g C

Moles H2O:

4.32g H2O * (1mol/18.01g) = 0.2399 moles H2O * (2mol H / 1molH2O) = 0.4797moles H = 0.4797g H

Mass O:

12.01g = Mass O + 3.8423g C + 0.4797g H

Mass O = 7.688g O

Moles O:

7.688g O * (1mol/16g) = 0.48 moles O

The ratio of atoms (Dividing in the moles of C that are the lower number of moles):

O: 0.48moles O / 0.3199 moles C = 1.50

C: 0.3199 moles C / 0.3199 moles C = 1

H: 0.4797 moles H / 0.3199 moles C = 1.50

As empirical formula requires whole numbers:

O: 1.50* 2 = 3

C: 1*2 = 2

H: 1.50*2 = 3

The empirical formula is:

C2H3O3

Which of the following is a physical change?

Answers

the awnser i think is c

If you ran the reaction for this experiment and began with 65.0 mmol of isopentyl alcohol, how many grams of isopentyl acetate could you theoretically produce assuming only a 77.0% attainable yield

Answers

Answer:

6.52g = Actual yield (g)

Explanation:

The yield of a reaction is:

Percent yield = Actual yield (g) / Theoretical Yield (g) * 100

As 1 mol of isopentyl alcohol produce 1 mol of isopentyl acetate (Theoretical Yield), the theroretical yield of isopentyl acetate is 65.0mmol = 0.0650mol. To solve this question we need to convert the moles of isopentyl acetate to mass using its molar mass (130.19g/mol).

With the equation of percent yield we can find the mass obtained as follows:

Theoretical yield:

0.0650mol * (130.19g/mol) = 8.46g of isopentyl alcohol

Mass produced:

77 = Actual yield (g) / 8.462g * 100

6.52g = Actual yield (g)

The mass of isopentyl acetate that can be produced is 6.52 g

Balanced equation

See attached photo

From the balanced equation,

1 mole of isopentyl alcohol reacted to produce 1 mole of isopentyl acetate.

Therefore,

65 mmole (i.e 0.065 mole) of isopentyl alcohol will also react to produce 0.065 mole of isopentyl acetate.

How to determine the actual yield (in mole) Percentage yield = 77%Theoretical yield = 0.065 mole Actual yield =?

Actual yield = percent × theoretical

Actual yield = 77% × 0.065

Actual yield = 0.05005 mole

How to determine the mass Mole of isopentyl acetate = 0.05005 mole Molar mass of isopentyl acetate = 130.19 g/molMass of isopentyl acetate =?

Mass = mole × molar mass

Mass of isopentyl acetate = 0.05005 × 130.19

Mass of isopentyl acetate = 6.52 g

Learn more about stoichiometry:

https://brainly.com/question/14735801

A sealed container was filled with 0.300mol H2(g), 0.400mol I2(g), and 0.200mol HI(g) at 870K and total pressure 1.00bar. Calculate the amounts of the components in the mixture at equilibrium given that K.= 70 for the reaction H2(g)+I2(g) --> 2HI(g).

Answers

Answer:

[HI] = 0.704mol

[H2] = 0.048mol

[I2] = 0.148mol

Explanation:

Based on the equilibrium:

H2(g)+I2(g) --> 2HI(g)

The equilibrium constant, K, is defined as:

K = 70 = [HI]² / [H2] [I2]

Where [] could be taken as the moles in equilibrium of each reactant

To know the direction of the equilibrium we need to find Q with the initial moles of each species:

Q = [0.200mol]² / [0.300mol] [0.400mol]

Q = 0.333

As Q < K, the reaction will shift to the right producing more HI. The equilibrium moles are:

[HI] = 0.200mol + 2X

[H2] = 0.300mol - X

[I2] = 0.400mol - X

Replacing in K:

70 = [0.200 + 2X ]² / [0.300 - X] [0.400 - X]

70 = 0.04 + 0.8 X + 4 X² / 0.12 - 0.7 X + X²

8.4 - 49 X + 70 X² = 0.04 + 0.8 X + 4 X²

8.36 - 49.8X + 66X² = 0

Solving for X:

X = 0.252 moles. Right solution

X = 0.502 moles. False solution. Produce negative moles.

Replacing:

[HI] = 0.200mol + 2*0.252 mol

[H2] = 0.300mol - 0.252 mol

[I2] = 0.400mol - 0.252 mol

[HI] = 0.704mol

[H2] = 0.048mol

[I2] = 0.148mol

I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine solution.

Required:
What is the pH after the addition of a total of 201 mL of the nitric acid?

Answers

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

Moles C2H5NH2:

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

Moles HNO3:

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

pH = 4.00 is the pH of the mixture

what is the qualitative analysis of (nh4) 2co3 using NaOH, HCL, BaCL2, and AgNO3​

Answers

Answer:

qualatatiev is fs-hj_jakakak

The action of some commercial drain cleaners is based on the following reaction:
2 NaOH(s) + 2 Al(s) + 6 H2O(l) â 2 NaAl(OH)4(s) + 3 H2(g)
What is the volume of H2 gas formed at STP when 4.32 g of Al reacts with excess NaOH?
A. 3.59 L
B. 2.39 L
C. 5.87 L
D. 5.38 L

Answers

Answer:

5.38 L

Option D.

Explanation:

2 NaOH(s) + 2 Al(s) + 6 H₂O(l)   →  2 NaAl(OH)₄(s)  + 3 H₂(g)

We convert mass of Al to moles:

4.32 g . 1 mol /26.98g = 0.160 moles

As NaOH is in excess, aluminum is the limiting reactant.

We see stoichiometry, were ratio is 2:3.

2 moles of Al can produce 3 moles of hydrogen

Our 0.160 moles may  produce (0.160 . 3)/2 = 0.240 moles of H₂.

We know that 1 mol of any gas at STP conditions is contained in 22.4L

So let's make the conversion factor:

0.240 mol . 22.4L / 1mol = 5.38 L

important of organic chemistry in our daily life ​

Answers

Answer:

Human life has become very simple by using organic chemicals. The importance of organic chemicals in the daily life and industrial area can be explained as follows. (i) Food Vitamins, proteins, sugar, flour, fats etc. ... (ii) Agriculture is an important place for organic chemical for the growth of agricultural production.

A sample of gas is placed into an enclosed cylinder and fitted with a movable piston. Calculate the work (in joules) done by the gas if it expands from 5.33 L to 11.05 L against a pressure of 1.50 atm.

Answers

Explanation:

here is the answer. Feel free to ask for more chem help

What occurs in endothermic reactions?
a) Heat energy is absorbed.
b) Water is produced.
c) Oxygen is produced.
d) Heat energy is released.

Answers

Answer:

a) Heat energy is absorbed.

Explanation:

What occurs in endothermic reactions?

a) Heat energy is absorbed. YES. This is the definition of an endothermic reaction.

b) Water is produced. NO. Endothermic refers to the heat absorbed and not to the products formed.

c) Oxygen is produced. NO. Endothermic refers to the heat absorbed and not to the products formed.

d) Heat energy is released. NO. This is known as an exothermic reaction.

Determine whether the stopcock should be completely open, partially open, or completely closed for each activity involved with titration.
Close to the calculated endpoint of a titration ________
At the beginning of a titration _______
Filling the buret with titrant ________
Conditioning the buret with titrant _______

Answers

Answer:

Close to the calculated endpoint of a titration - Partially open

At the beginning of a titration - Completely open

Filling the buret with titrant - Completely closed

Conditioning the buret with the titrant - Completely closed

Explanation:

'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.

As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.

The elemental mass percent composition of succinic acid is 40.68% CC, 5.12% HH, and 54.19% OO. Determine the empirical formula of succinic acid.

Answers

Answer:

C2H3O2

Explanation:

Empirical formula is defined as the simplest whole-number ratio of atoms present in a molecule. To solve this question we need to convert the percentage of each atom to moles using molar mass. With the moles of each atom we can find the ratio:

Moles C -Molar mass: 12.01g/mol-

40.68g * (1mol/12.01g) = 3.387 moles C

Moles H-Molar mass:1g/mol-:

5.12g * (1mol/1g) = 5.12 moles H

Moles O -Molar mass: 16g/mol-

54.19g * (1mol/16g) = 3.387 moles O

Ratio of atoms (Dividing in moles of C that are the lower number of moles):

C = 3.387 moles C / 3.387moles C = 1

H = 5.12moles H / 3.387moles C = 1.5

O = 3.387moles O / 3.387 moles C = 1

This ratio twice (To have only whole-numbers):

C = 2

H = 3

O = 2

Empirical formula of succinic acid:

C2H3O2

The angular momentum quantum number (l) value of 0 indicates the ________ subshell.

Answers

Answer:

indicates the number of subshells

Question 65 pts
(07.02 MC)

During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)

Group of answer choices

It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.

It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.

It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.

It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.

Answers

Answer:

It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed

Explanation:

A reaction may be endothermic or exothermic. In an endothermic reaction, energy is absorbed by the process while in an exothermic process energy is given out by the process.

Recall that the enthalpy change of a reaction = enthalpy of products - enthalpy of reactants

Hence, where the energy required to break bonds in the reactants is less than the energy released when the products are formed, the reaction is endothermic.

For an endothermic reaction, the enthalpy change of the reaction is positive.

In this case, enthalpy of reaction = 920 - (-750) = 1670 kJ/mol

the force of attraction between non polar molecules are what​

Answers

Answer:

dispersion force

Explanation:

it’s dispersion force bro

What is the sum of the coefficients of the balanced equation for the following reaction: FeCl2(aq) K2Cr2O7(aq) HCl(aq) ---> CrCl3(aq) FeCl3(aq) KCl(aq) H2O(l)

Answers

Answer:

The unbalanced chemical equation is

K

2

Cr

2

O

7

+HCl→KCl+CrCl

3

+H

2

O+Cl

2

Balance all atoms except H and O.

K

2

Cr

2

O

7

+10HCl→2KCl+2CrCl

3

+H

2

O+Cl

2

Assign oxidation numbers.

K

2

+6

Cr

2

O

7

+10H

−1

Cl

→2KCl+2

+3

Cr

Cl

3

+H

2

O+

0

Cl

2

The oxidation number of Cr decreases from +6 to +3. Total decrease in the oxidation number of two Cr atoms is 6. The oxidation number of Cl increases from -1 to 0. Total increase in the oxidation number of 2 Cl atoms is 2.

Balance increase in oxidation number with decrease in oxidation number by using appropriate coefficients.

K

2

Cr

2

O

7

+14HCl→2KCl+2CrCl

3

+H

2

O+3Cl

2

Balance O atoms by adding six water molecules on products side.

K

2

Cr

2

O

7

+14HCl→2KCl+2CrCl

3

+7H

2

O+3Cl

2

H atoms are balanced.

K

2

Cr

2

O

7

+14HCl→2KCl+2CrCl

3

+7H

2

O+3Cl

2

This is balanced chemical equation.

The sum of the coefficients of the products is 2+2+7+3=14

What is the density of Ar(g) at -11°C and 675 mmHg?

Answers

Answer:

The Density Of Ar (g) At -11°C And 675 MmHg (R =0.08206 L·atm/mol·K, 1 Atm = 760mmHg).

What does the "R-" represent?

A. a halogen
B. a hydroxyl group
C. an alkyl group
D. an oxyacid


URGENT!! ​

Answers

Answer:

general formula RCOX, where R represents an alkyl or aryl organic radical group, CO ... represents a halogen atom such as chlorine ... loss of a hydroxyl group (-OH), viz, acetyl,. CH, CO- ..

The data shows the number of years that 30 employees worked for an insurance company before retirement. is the population mean for the number of years worked, and % of the employees worked for the company for at least 10 years. (Round off your answers to the nearest integer.)

Answers

Answer:

14

73%

Explanation:

The mean Number of years worked :

. (sum of service years) / employees in the

(8+13+15+3+13+28+4+12+4+26+29+3+10+3+17+13+15+15+23+13+12+1+14+14+17+16+7+27+18+24) /

(417 / 30)

= 13.9 years

= 14 years

The percentage of employees who have worked for atleast 10 years :

Number of employees with service years ≥ 10 years = 22 employees

Total number of employees

Percentage (%) = (22 / 30= * 100% = 0.7333 * 100% = 73.33% = 73%

What effect does a high carbon level have on a deep ocean

Answers

Explanation:

High carbon concentration in the deep ocean means increased absorption of carbon to the atmosphere resulting to even greater and harmful amounts of carbon in the atmosphere. Therefore we need to keep a close eye of the deep ocean in the quest to monitor and pump out excess carbon from this part of marine life.

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