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Some chemical reactions that are reversible (put some picture for better understanding)

GalinKa [24]

Ammonium chloride is a white solid that breaks down when heated and produces ammonia and hydrogen chloride.

<h3>What is the reversible reaction?</h3>

A reversible reaction is a reaction in which the conversion of reactants to products and products to reactants occur at the same time. In the above example, the chemical shows a reversible reaction because it moves both forward and backward direction. In reversible reaction, equal amount of reactant is converted into product and product into reactant.

So we can conclude that Ammonium chloride is a chemical that represents a reversible reaction.

Learn more about reaction here: brainly.com/question/11231920

#SPJ1

8 0

2 years ago

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

S = $K_{b} \times ln \Omega$

where, S = entropy

$K_{b}$ = Boltzmann constant

$\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

$\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

= $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

= $1.38 \times 10^{-23} \times (-0.182)$

= $-0.251 \times 10^{-23}$

or, = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

8 0

3 years ago

The central atom in ________ does not violate the octet rule. the central atom in ________ does not violate the octet rule. sf4

ruslelena [56]

We have to know which do not violet octet rule.

The central atom in SF₄ does not violate the octet rule. the central atom in CF₄ does not violate the octet rule.

As per octet rule, atoms can combine either by transfer of valence electrons from one atom to another or by sharing of valence electrons in order to have an octet in their valence shell.

The central atom S of SF₄ contains 8 electrons in the valence shell. Also, C atom of CF₄ contains 8 electrons in the valence shell.

ICl₄⁻ contains 9 electrons in the outermost shell.

7 0

3 years ago

If a neutral atom of boron (B) contains 5 protons, 5 electrons and 6 neutrons, what is the mass number of this atom? (2 points)

Shkiper50 [21]

I believe the answer is 11, its been awhile since i was in science

6 0

3 years ago

Read 2 more answers

NEED HELP ASAP WITH THESE QUESTIONS GIVING FAIR AMOUNT OF POINTS IF HELPED WITH ALL QUESTIONS Violet light has a wavelength of 4

scoray [572]

Answer:

a) 7.14e19 Hz

b) 2.298e-27 J

c) 2.793e-19 J; 7.117e9 nm

d) 7.5e14 Hz; 4.96e-19 J

e) 6.2947e-18 J; 31.6 nm

f) 2.21e-22 J

g) 7.1e-19 J; 1.1e15 Hz

h) 3.422e-19 J; 581 nm

i) 4.2e14 Hz

j) 1.92e8 m

k) 7.14e16 Hz; Ultraviolet

Explanation:

Frequency: ν Wavelength: λ Energy: E Speed of light: C (3.00e8) Planck's Constant: h (6.626e-34)

ν -> λ λ = C/ν

λ -> ν ν = C/λ

For either of these equations, wavelength must be converted to meters or nanometers, depending on the equation.

For ν -> λ, after doing the equation, convert the wavelength into nanometers by dividing by 1e-9.

For converting λ -> ν, convert the wavelength into meters by multiplying by 1e-9.

For energy: E = hν = hc/λ

Now that the setup is out of the way:

a) Violet light has a wavelength of 4.20 x 10-12 m. What is the frequency?

λ -> ν ν = C/λ

$\frac{3.00e8}{4.20e-12}$ = 7.14e19 Hz

b) A photon has a frequency (n) of 3.468 x 106 Hz. Calculate its energy

E = hν = hc/λ

(6.626e-34) (3.468e6) = 2.298e-27 J

c) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 4.215 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (4.215e14) = 2.793e-19 J

ν -> λ λ = C/ν

$\frac{3.00e8}{4.215e14}$ = 7.117 m

$\frac{7.117m}{1}*\frac{1nm}{1e-9m}$ = 7.117e9 nm

d) Calculate the frequency and the energy of blue light that has a wavelength of 400 nm (h = 6.62 x 10-34 J-s).

λ -> ν ν = C/λ

$\frac{400 nm}{1} *\frac{1e-9m}{1nm}$ = 4e-7 m

$\frac{3.00e8}{4e-7}$ = 7.5e14 Hz

E = hν = hc/λ

(6.626e-34) (7.5e14) = 4.96e-19 J

e) Calculate the wavelength and energy of light that has a frequency of 9.5 x 1015 Hz.

ν -> λ λ = C/ν

$\frac{3.00e8}{9.5e15}$ = 3.16e-8 m

$\frac{3.16e-8m}{1}*\frac{1nm}{1e-9m}$ = 31.6 nm

E = hν = hc/λ

(6.626e-34) (9.5e15) = 6.2947e-18 J

f) A photon of light has a wavelength of 0.090 cm. Calculate its energy.

E = hν = hc/λ

Convert the wavelength from cm to meters:

$\frac{0.090cm}{1} *\frac{1m}{100cm}$ = 9e-4 m

$\frac{(6.626e-34)(3.00e8)}{9e-4}$ = 2.21e-22 J

g) Calculate the energy and frequency of red light having a wavelength of 2.80 x 10-5 cm.

E = hν = hc/λ

Convert the wavelength from cm to meters:

$\frac{2.80e-5cm}{1} *\frac{1m}{100cm}$ = 2.8e-7 m

$\frac{(6.626e-34)(3.00e8)}{2.8e-7}$ = 7.1e-19 J

λ -> ν ν = C/λ

Convert the wavelength from cm to meters:

$\frac{2.80e-5cm}{1} *\frac{1m}{100cm}$ = 2.8e-7 m

$\frac{3.00e8}{2.8e-7}$ = 1.1e15 Hz

h) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 5.165 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (5.165e14) = 3.422e-19 J

ν -> λ λ = C/ν

$\frac{3.00e8}{5.165e14}$ = 5.81e-7 m

$\frac{5.81e-7m}{1}*\frac{1nm}{1e-9m}$ = 581 nm

i) The wavelength of green light from a traffic signal is centered at 7.20 x 10-5 cm. Calculate the frequency.

λ -> ν ν = C/λ

Convert the wavelength from cm to meters:

$\frac{7.20e-5 cm}{1} *\frac{1m}{100cm}$ = 7.2e-7 m

$\frac{3.00e8}{7.2e-7}$ = 4.2e14 Hz

j) If it takes 1.56 seconds for radio waves (which travel at the speed of light) to reach the moon from Earth, how far away is the moon?

All we want to do here is to convert frequency (speed) to wavelength (distance). This problem requires a bit of thought, but it isn't bad once you realize that frquency is speed and wavelength is distance. It becomes just like the other problems after that. Also, I'll leave this distance in meters, but I think you can figure out how to convert it if it wants it in another unit.

One second is equal to 1 Hertz, so our frequency is 1.56 Hz.

ν -> λ λ = C/ν

$\frac{3.00e8}{1.56}$ = 1.92e8 m

The actual distance from the earth to the moon via google is 3.84e7, but sometimes problems like this will mess with the numbers to make sure that you didn't just look up the answer. I'm still pretty sure that this is right, however.

k) Calculate the frequency of light that has a wavelength of 4.20 x 10-9m. Identify the type of electromagnetic radiation.

First, we convert wavelength to frequency, as normal:

λ -> ν ν = C/λ

$\frac{3.00e8}{4.20e-9}$ = 7.14e16 Hz

Then we identify the electromagnetic wave type. You can look up a conversion chart for these on google, but since our frequency is in the e15 - e17 range, this light is considered ultraviolet.

5 0

4 years ago