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erastovalidia [21]
2 years ago
10

Given that ethylene has a λmax of 175nm, butadiene has a λmax of 220nm, and 2-methyl-1,3-butadiene has a λmax or 215nm, what is

the λmax of 2,3,4-trimethylhexatriene?
Physics
1 answer:
Vika [28.1K]2 years ago
5 0

The λmax of 2,3,4-trimethylhexatriene is 280 nm.

Ethylene has a λmax of 175nm.

Butadiene has a λmax of 220nm.

2-methyl-1,3-butadiene has a λmax or 215nm.

1,3,5-hexatriene has a λmax of 258nm.

Woodward's rules, sometimes known as Woodward-Fieser rules (after Louis Fieser) and named after Robert Burns Woodward, are a number of sets of empirically developed principles that aim to forecast the wavelength of the absorption maximum (max) in an ultraviolet-visible spectrum of a certain molecule.

By using the Woodward Fieser rule,

R- (Alkyl Group) .... +5 nm = 5 × 2 = 10

RO- (Alkoxy Group) .. +6 = 6 × 2 = 12

Adding 22nm to the λmax of 1,3,5-hexatriene as it has 2 alkyl groups and 2 alkoxy groups to form 2,3,4-trimethylhexatriene.

The λmax of 2,3,4-trimethylhexatriene is 280 nm.

Learn more about Woodward-Fieser here:

brainly.com/question/16982345

#SPJ4

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6 0
2 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

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Answer:

It will take 5 seconds

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choose the correct answer 14: which of the following force follows the inverse square law of distance A: gravitaional force B: e
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Answer:

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