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gizmo_the_mogwai [7]
3 years ago
7

A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo

w, sending the light through a diffraction grating, and measuring the positions of first-order spectral lines on a detector 15.0 cm behind the grating. Unfortunately, she has lost the card that gives the specifications of the grating. Fortunately, she has a known compound that she can use to calibrate the grating. She heats the known compound, which emits light at a wavelength of 501 nm, and observes a spectral line 9.95 cm from the center of the diffraction pattern. PART A:
What is the wavelength emitted by compound A that have spectral line detected at position 8.55 cm?


PART B:


What is the wavelength emitted by compound B that have spectral line detected at position and 12.15 cm?
Physics
1 answer:
padilas [110]3 years ago
4 0

Answer:

 a)    λ = 189.43 10⁻⁹ m  b)    λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

      d sin θ= m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

        tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

      tan θ = sin θ / cos θ = sin θ

We replace

      d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

     d = λ  L / y

     d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

     d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

     λ  = d y / L

    λ  = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

    λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

      λ  = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

      λ = 269.19 10⁻⁹ m

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1.92 kg of nitrogen.

Explanation:

The following data were obtained from the question:

Heat absorbed (Q) = 384000 J

Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol

Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.

This is illustrated below:

Q = mol·ΔHv

384000 = mole of N2 x 5600

Divide both side by 5600

Mole of N2 = 384000/5600

Mole of N2 = 68.57 moles

Next, we shall convert 68.57 moles of nitrogen, N2 to grams.

This can be obtained as follow:

Molar mass of N2 = 2 x 14 = 28 g/mol.

Mole of N2 = 68.57 moles.

Mass of N2 =..?

Mole = mass /molar mass

68.57 = mass of N2 /28

Cross multiply

Mass of N2 = 68.57 x 28

Mass of N2 = 1919.96 g

Finally, we shall convert 1919.96 g to kilograms.

This can be achieved as shown below:

1000g = 1 kg

Therefore,

1919.96 g = 1919.96/1000 = 1.92 kg.

Therefore, 1.92 kg of nitrogen were burned off.

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