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gulaghasi [49]
3 years ago
11

Give the relationship(s) for any pair of protons with the proper term(s). Label – your choice. A.Heterotopic B.Heterotopic, dias

tereotopic C.Homotopic D.Homotopic, enantiotopic

Physics
1 answer:
Afina-wow [57]3 years ago
7 0

Answer and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

Hope this helps!!!

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A bowler throws a bowling ball of radius R = 11.0 cm down the lane with initial speed = 8.50 m/s. The ball is thrown in such a w
Ad libitum [116K]

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

<u>Step 1: </u>Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

<em></em>

<em>(a) For what length of time does the ball skid?</em>

The velocity at time t can be written as v(t) = v0 + at

 ⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

   ⇒The acceleration (negative) due to friction: a = -µg

           ⇒ with µ = 0.210

          ⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t  = 46.9*0.11*t = 5.159t

7.219t = 8.5

<u>t = 1.18 seconds</u>

<em>b) How far down the lane does it skid?</em>

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = <u>8.6 m</u>

<em>c) How many revolutions does it make before it starts to roll?</em>

The angular acceleration of the ball is:

α =  τ/I

 ⇒ with  τ = the torque experienced by the ball due the frictional force

   ⇒  τ = fk*R

α = fk*R /I

 ⇒ I = 2/5 m*R²

 ⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

<u>Number of revolutions = 5.19</u>

<em>(d) How fast is it moving when it starts to roll?</em>

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = <u>6.07 m/s</u>

7 0
3 years ago
Unpolarized light is passed through three successive polaroid filters, each with its transmission axis at 32. 8° to the precedin
Oliga [24]

The percentage of light gets through three successive polarized filters is 24.9 %

From the question,

Given that,

Angle of transmission axis = 32.8°

The intensity of light emerging from the first polarizer is determined by the equation

I₁ = I₀ / 2

where I₀ ⇒ intensity of unpolarized light

The light emerging from the second polarizer can be mathematically represented by,

I₂ = I₁ x cos²θ

Substituting the values,

I₂ =( I₀/2) x cos²θ

  = (I₀/2) x cos² (32.8 )

  = (I₀/2) x 0.706

  = (0.706 / 2 ) x I₀

The light emerging from the third polarizer is represented as,

I₃ = I₂ x cos²θ

Substituting the values in the above equation,

I₃ = ( 0.706 / 2 ) I₀ x cos² (32.8)

  = (0.706 / 2 ) I₀ x 0.706

  = 0.249 I₀

The percentage of intensity of light  that gwts through with respect to the intensity of unpolarized light is given by the equation,

(I₃ / I₀) x 100

Substituting the values

[(0.249 x I₀) / I₀ ] x 100 = 24.9 %

Hence the percentage of light gets through is 24.9%

To learn more about unpolarized light: brainly.com/question/17164167

#SPJ4

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2 years ago
There are four springs stretched by the same mass.
kirill115 [55]

Answer:

That would be Spring C

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i took the test :P

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Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom
Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

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B = B_o(-\hat k)

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v = v_o \hat j

now the force on the moving charge is given as

\vec F = q(\vec v \times \vec B)

now we have

\vec F = (-e)(v_o \hat j \times B_o(-\hat k))

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A rising pendulum bob gains _____ energy.
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