Chemistry: Flame Test Questions And Answers
Part One (Flame Test)
1. Create and complete a data table for Part One of the lab. It should include the name of the element (or unknown) examined and the color of the observed flame.
Solutions
Names
Known 1
Barium
Known 2
Calcium
Known 3
Sodium
Known 4
Rubidium
Known 5
Potassium
Known 6
Lithium
Unknown 1
Sodium
Unknown 2
Potassium
2. Identify each unknown from Part One of the lab and briefly explain why you identified each unknown as you did.
The first unknown from part one from this virtual lab is Sodium, as the flames colors were burning with the same exact flame as the example Sodium. The second unknown from part one from this virtual lab is Potassium, the flames remained constant with the conclusions from the potassium.
Part Two (Spectroscopy) …show more content…
Describe the line spectrum of the star. Give the color and wavelength value of the five brightest lines in the spectrum.
This line spectrum is extending from the color purple meaning that the purple is first to red meaning that the reddest is the last, and a lot of other colors in the line spectrum. The wavelengths and the colors of the five brightest lines are the sentences listed below.
1. Color = Purple, wavelength which equals 415 nm.
2. Color = Teal, wavelength which equals 490 nm.
3. Color = Green, wavelength which equals 550 nm.
4. Color = Yellow, wavelength which equals 585 nm.
5. Color = Red, wavelength which equals 710 nm.
Conclusion
1. These dissimilar elements created dissimilar colors of the light when they were heated because there are a lot of dissimilar amounts of electrons, and they’re all activated to a dissimilar degree. These electrons diffuse light because of the kinetic energy that is being freed when coming back to their original start area. The properties that alter the colors when the light is being created is also the distance and the number of the electrons.
2. These elements must be heated before they diffused colored light because they can stretch to the outermost electron ring that results in this arriving at the greatest amount of kinetic
1. Create and complete a data table for Part One of the lab. It should include the name of the element (or unknown) examined and the color of the observed flame.
Solutions
Names
Known 1
Barium
Known 2
Calcium
Known 3
Sodium
Known 4
Rubidium
Known 5
Potassium
Known 6
Lithium
Unknown 1
Sodium
Unknown 2
Potassium
2. Identify each unknown from Part One of the lab and briefly explain why you identified each unknown as you did.
The first unknown from part one from this virtual lab is Sodium, as the flames colors were burning with the same exact flame as the example Sodium. The second unknown from part one from this virtual lab is Potassium, the flames remained constant with the conclusions from the potassium.
Part Two (Spectroscopy) …show more content…
Describe the line spectrum of the star. Give the color and wavelength value of the five brightest lines in the spectrum.
This line spectrum is extending from the color purple meaning that the purple is first to red meaning that the reddest is the last, and a lot of other colors in the line spectrum. The wavelengths and the colors of the five brightest lines are the sentences listed below.
1. Color = Purple, wavelength which equals 415 nm.
2. Color = Teal, wavelength which equals 490 nm.
3. Color = Green, wavelength which equals 550 nm.
4. Color = Yellow, wavelength which equals 585 nm.
5. Color = Red, wavelength which equals 710 nm.
Conclusion
1. These dissimilar elements created dissimilar colors of the light when they were heated because there are a lot of dissimilar amounts of electrons, and they’re all activated to a dissimilar degree. These electrons diffuse light because of the kinetic energy that is being freed when coming back to their original start area. The properties that alter the colors when the light is being created is also the distance and the number of the electrons.
2. These elements must be heated before they diffused colored light because they can stretch to the outermost electron ring that results in this arriving at the greatest amount of kinetic