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2 years ago

The c-cl bond dissociation energy in cf3cl is 339 kj/mol. What is the maximum wavelength of photons that can rupture this bond?

Physics

2 answers:

Rom4ik [11]2 years ago

8 0

Answer:

3.53*10^-7 color white "m

Where h= Planck constant

C=speed of light

Explanation:

Calculating energy required to dissociate the cl--cl Bond as this

Energy= +339*color white*kJmol

E=(339*10^3/6.022*10^23)=5.63*10^-19color white

But note from Planck equation

E=hf=h*c/lamda

Lamda= hc/E

=6.63*10^-34*3*10^8/5.63*10^-19

Lamda=3.532*10^-7 colorwhite*m

ruslelena [56]2 years ago

7 0

Answer:

3.53*10^{-7} m

Explanation:

Photon that can rupture the bonds are those with the energy of the bond dissociation energy. If we want to know the energy for each molecule we have to take into account that:

$1mol=6.022*10^{23}molecule$

Hence, we have

$E_d=339\frac{10^{3}J}{mol}*\frac{1mol}{6-022*10^{23}molecules}=5.62*10^{-19}J/molecule$

but the energy is also:

$E_d=h\nu =\frac{hc}{\lambda}\\\\\lambda=\frac{hc}{E_d}$

where h is the Planck's constant and c is the speed of ligth. By replacing we obtain:

$\lambda=\frac{(6.62*10^{-34}Js)(3*10^{8}m/s)}{5.62*10^{-19}J}=3.53*10^{-7}m$

hope this helps!

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Break down the speeds in two moments

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vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

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3 years ago

Is there a frame of reference one can go into that seems to eliminate gravity as Newton described it?

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Answer:

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2 years ago

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

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Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

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$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

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Answer:

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