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What is the approximate wavelength of light emitted by each of the following?

1. **C (yellow):**
- a. 350 nm
- b. 425 nm
- c. 590 nm
- d. 625 nm

2. **CaCl\(_2\) (orange):**
- a. 350 nm
- b. 425 nm
- c. 590 nm
- d. 625 nm

3. **KCl (reddish-violet):**
- a. 350 nm
- b. 425 nm
- c. 590 nm
- d. 625 nm

4. **NaCl (yellow):**
- a. 350 nm
- b. 425 nm
- c. 590 nm
- d. 625 nm

5. **LiCl (reddish-violet):**
- a. 350 nm
- b. 425 nm
- c. 590 nm
- d. 625 nm

Answer :

Final answer:

The wavelengths of the light emitted when various compounds are burned can be approximately identified as follows: Carbon (C) and Sodium Chloride (NaCl) emit yellow light at 590 nm, Calcium Chloride (CaCl2) emits orange light also at 590 nm, while Potassium Chloride (KCl) and Lithium Chloride (LiCl) emit reddish-violet light at 425 nm.

Explanation:

The approximate wavelength of light emitted by different elements and compounds (flame test colors) correspond to a specific spectrum of visible light. For your specific questions, these would be the answers:

  • For Carbon (C) which produces a yellow flame, the wavelength is approximately 590 nm.
  • Calcium Chloride (CaCl2) produces an orange flame, which has an approximate wavelength of 590 nm.
  • Potassium Chloride (KCl) emits a reddish-violet flame, this has an approximate wavelength of 425 nm.
  • Sodium Chloride (NaCl) also yields a yellow flame which corresponds to a wavelength of roughly 590 nm.
  • Lastly, Lithium Chloride (LiCl) gives a reddish-violet flame with a wavelength of approximately 425 nm.

Learn more about Wavelength of Emitted Light here:

https://brainly.com/question/34403004

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Rewritten by : Jeany

yes I pretty sure it's c

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