Predict the changes in the number of particles based the diagram.A) Pressure is indirectly related to moles.
B) Volume is proportionally related to moles.
C) Moles remains constant as pressure changes.
D) Temperature is directly related to the moles.

Answer:

1.74845

Explanation:

We have the following reaction:

I2 + H2 => 2 HI

Now, the constant Kc, has the following formula:

Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b

In this case I2 is A, H2 is B and C is HI

We know that the values are:

 H2 = 1 × 10 ^ -3 at 448 ° C

I2 = 2 × 10 ^ -3 at 448 ° C

HI = 1.87 × 10 ^ -3 at 448 ° C

Replacing:

Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}

Kc = 1.87 ^ 2/2 * 1

Kc = 1.74845

Which means that at 448 ° C, Kc is equal to 1.74845

Answer:

Explanation:

[H2] = 10^-3

[I2] = 2*10^-3

[HI] = 0

in equilbiirum

[H2] = 10^-3 - x

[I2] = 2*10^-3 -x

[HI] = 0 + 2x

and we know

[HI] = 0 + 2x = 1.87*10^-3

x = ( 1.87*10^-3)/2 =  0.000935

then

[H2] = 10^-3 - 0.000935 = 0.000065

[I2] = 2*10^-3 -0.000935 = 0.001065

                        H₂                +          I              ⇄              2 HI

Initially     1 × 10⁻³                       2 × 10⁻³

Change  -9.35 × 10⁻⁴               -9.35 × 10⁻⁴                +1.87 × 10⁻³

At equil   6.5 × 10⁻⁵                  1.06 5 × 10⁻³               1.87 × 10⁻³

HI increase by 1.87 × 10⁻³M

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

, where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

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