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Serjik [45]
2 years ago
12

2) Of the following, which has the shortest de Broglie wavelength? A) an airplane moving at a velocity of 300 mph B) a helium nu

cleus moving at a velocity of 1000 mph C) a nitrogen molecule moving at a velocity of 1000 mph D) a nitrogen molecule moving at a velocity of 5000 mph
Chemistry
1 answer:
8090 [49]2 years ago
5 0

Answer:

B) a helium nucleus moving at a velocity of 1000 mph

Explanation:

According to the De Broglie relation

λ= h/mv

h= planks constant

m= mass of the body

v= velocity of the body.

As we can see from De Broglie's relation, the wavelength of matter waves depends on its mass and velocity. Hence, a very small mass moving at a very high velocity will have the greatest De Broglie wavelength.

Of all the options given, helium is the smallest matter. A velocity of 1000mph is quite high hence it will have the greatest De Broglie wavelength.

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If the composition of the reaction mixture at 400 k is [brcl] = 0.00415 m, [br2] = 0.00366 m, and [cl2] = 0.000672 m, what is th
Scorpion4ik [409]
Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl

According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]

and we have [Br2] = 0.00366 m  [Cl2]= 0.000672 m  [BrCl] = 0.00415 m

by substitution:
                          = [0.00415]^2 / ( [0.00366] * [0.000672])
             ∴   Q   = 7
7 0
2 years ago
Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)
marissa [1.9K]

Answer:

The\Delta H_{rxn} of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)

Enthalpy of each reactant and products are as follows.

\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}

\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}

\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}

\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}

=-347.4-2\times416.2+615.0+435.1

=-129.6 \,kJ

Therefore, The\Delta H_{rxn} of the given reaction is -129.6 kJ

4 0
1 year ago
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