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erastovalidia [21]
1 year 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]1 year 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

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By what factor would your weight increase if you could stand on the sun? (never mind that you can't.)
Fofino [41]

Gravity on the surface of sun is given as

g = \frac{GM}{R^2}

here we know that

M = 1.98 \times 10^{30} kg

R = 6.95 \times 10^8 m

now we will have

g = \frac{(6.67 \times 10^{-11})(1.98 \times 10^30)}{(6.95 \times 10^8)^2}

g = 273.4 m/s^2

now we need to find the ratio of weight on surface of sun and on surface of Earth

R = \frac{mg_{sun}}{mg_{earth}}

R = \frac{273.4}{9.8} = 28 times

so weight will increase by 28 times

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A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J
cestrela7 [59]

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = \frac{energy}{time}

The energy needed to light one bulb=power*time

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Energy = 32W*60sec=1920J

Therefore energy needed to light one bulb is 1920J

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An object is located 70 cm from a concave mirror with a focal length of 15 cm. What is the image
Stels [109]

(a) The distance of the image formed by the concave mirror is 19.1 cm.

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<h3>Image distance </h3>

The distance of the image formed by the concave mirror is calculated as follows;

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/15 - 1/70

1/v = 0.05238

v = 1/0.05238

v = 19.1 cm

The image distance is smaller than object distance, thus the image formed is diminished and real.

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1 year ago
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Ugo [173]

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5 0
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