All categories

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!

You might be interested in

Write the answer: physics ... i need help

Mariana [72]

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

7 0

3 years ago

A vertically polarized beam of light of intensity 100 W/m2 passes through two ideal polarizers. The transmission axis of the fir

TEA [102]

To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I=I_0 cos^2\theta$

Where,

$I_ {0}$ indicates the intensity of the light before passing through the polarizer,

I is the resulting intensity, and

$\theta$ indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

Since we have two objects the law would be,

$I=I_0cos^2\theta_1*cos^2(\theta_2-\theta_1)$

Replacing the values,

$I=100*cos^2(20)*cos^2(40-20)$

$I=100*cos^4(20)$

$I=77.91W/m^2$

Therefore the intesity of the light after it has passes through both polarizers is $77.91W/m^2$

7 0

3 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

3 years ago

A wave has a wavelength of 10 mm and a frequency of 14 hertz. What is its speed?

Katena32 [7]

0.14 meters per second .

3 0

3 years ago

A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a

olganol [36]

This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
 U = (1/2)kx^2
 U = (1/2)(5.3)(3.62-2.60)^2
 U = 2.75706 J

6 0

3 years ago