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maria [59]
3 years ago
14

t the R group in the reagent over the arrow be isopropyl. (i.e. The reagent is LiN[CH(CH3)2]2.) Now, fill in the blanks. This re

action will follow ______ rule and will form the _________ substituted alkene. a. Zaitzev's; more b. Zaitzev's; less c. Hofmann's; less d. Hofmann's; more
Chemistry
2 answers:
vladimir2022 [97]3 years ago
6 0

Answer:. A

Explanation:

T the R group in the reagent over the arrow be isopropyl. (i.e. The reagent is LiN[CH(CH3)2]2.) Now, fill in the blanks. This reaction will follow Zaitzev'srule and will form the more substituted alkene.

Tanzania [10]3 years ago
3 0

Answer:c. Hofmann's; less

Explanation:

The Hofmann elimination process, named after its discoverer, the German chemist August Wilhelm Von Hofmann rule states that the major alkene product is the <u>least substituted </u>and least stable product when it comes to asymmetrical amines.  The Hofmann elimination can be illustrated in t e formation of “Hoffman” products in elimination reactions  using a less common, LDA LiN[CH(CH3)2]2  Lithium Di-isopropyl Amide (LDA) usually with base  potassium t-butoxide  which will give rise to a less substituted alkene.

This reaction will follow Hofmann's_____ rule and will form the ___less______ substituted alkene.

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A laser used to read CDs emits red light of wavelength 700 nm. How many photons does it emit each second if its power is?(a) 0.1
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power = work/time

watt = joule/sec

a) power = 0.10 watt = 0.1joules/sec

work = power * time

= 0.10 * 1 =0.1joules

work is northing but energy

therefore energy is 0.1joules

according to planks quantum theory E = nhν where nis the no. of photons ; h is planks constant; ν isfrequency

hence, n = E/hν

or n = Eλ/hc (ν = c/λ)

n = (0.1*700*10^-9)/6.625*10^-34*3*10^8 = 0.3522*10^18photons

b) similar to the above calculation

here E = 1.0joules

n = (1.0*700*10^-9)/6.625*10^-34*3*10^8 = 0.3522*10^17photons

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3 years ago
The Rio Grande river flows through the South Texas Plains ecoregion of Texas. The river is a major source of water for cities an
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Answer:

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3 years ago
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What mass of carbon dioxide is produced from the complete combustion of 5.30x10-3 g of methane?
pshichka [43]

Answer:

1.45 x 10⁻² g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

5.30 x 10⁻³ g CH₄          1 mole              1 mole CO₂          44.007 g
---------------------------  x  ----------------  x  ---------------------  x  -----------------  =
                                      16.043 g            1 mole CH₄            1 mole

=  0.0145 g CO₂

=  1.45 x 10⁻² g CO₂

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