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N76 [4]
3 years ago
9

According to de Broglie, which of these objects does not have a wavelength?

Chemistry
1 answer:
polet [3.4K]3 years ago
7 0

Answer: Option D is correct.

Explanation: Equation given by de Broglie is:

\lambda =\frac{h}{mv}

where, \lambda = wavelength of the particle

h = Planck's constant

m = mass of the particle

v = velocity of the particle

In option A, football will have some mass and is moving with a velocity of 25 m/s, hence it will have some wavelength.

In Option B, unladen swallow also have some mass and is moving with a velocity of 38 km/hr, hence it will also have some wavelength.

In Option C, a person has some mass and is running with a velocity of 7 m/hr, hence it will also have some wavelength.

As, the momentum of these particles are large, therefore the wavelength will be of small magnitude and hence, is not observable.

From the above, it is clearly visible that all the options are having some wavelength, so option D is correct.

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eimsori [14]
2 boxes of A
Because C = A + B
2 of A = 20 grams
at the other hand we have 2 of B = 10
So 20 + 10 = 30 grams
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4 years ago
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Answer:

c.the speed of the bail decrease over time

6 0
3 years ago
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What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hy
Elanso [62]

Answer:

The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.

Explanation:

The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.

A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.

Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.

SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.

7 0
3 years ago
Suppose you have a calorimeter that contains 100.0 grams of water at an initial temperature of 25*C. A salt (2.19 g, 0.020 moles
katen-ka-za [31]
ΔH=MCΔT
ΔH=100 x 4.2 x 4
ΔH=1680

ΔH per mole = ΔH ÷ moles
ΔH per mole = 1680 ÷ 0.02
<span>ΔH per mole= 84000Jmol
</span>84000 ÷ 1000 = 84KJmol

its exothermic as heat is given out into the solution 
7 0
4 years ago
Read 2 more answers
The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. The balanced chemical reaction equatio
Bogdan [553]

Answer:

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

                C₁₂H₂₂O₁₁  +  12 O₂  ⇒   12 CO₂  +  11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

H° H₂O = -285.8 kJ/mol

Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

6 0
3 years ago
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