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Readme [11.4K]
3 years ago
13

Suppose the filament in a super-powerful flashlight is heated up to 3600 K. What type of light would be given off by this filame

nt, mostly?
Chemistry
2 answers:
Ksenya-84 [330]3 years ago
7 0

Answer:

The correct answer is "Infrared light".

Explanation:

Infrared light, also known as infrared radiation, is a type of light that can no be seen but it can be felt from the heat that it generates. The spectrum of wavelengths that comprise the infrared light is given from 700 nanometers, which is the edge of visible red light, up to 1 millimeter of wavelength. Near-Infrared light is the one that can reach the highest temperature, reaching from 3000 to 5200 K. Therefore, a super-powerful flashlight heated up to 3600 K can be given off infrared light.

alexdok [17]3 years ago
4 0
A cool white is the spectrum that lightbulbs between 3600-5000k would mainly show

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Answer:

a. 5.77L

b. 700mmHg

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Explanation:

Using PV = nRT we can solve these problems where:

P is pressure of the gas in atm (1atm = 760mmHg)

V is volume in liters

n are moles of the gas

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a. PV = nRT

V = nRT/P

P = 773mmHg*(1atm/760mmHg) = 1.017atm

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V = 0.240mol*0.082atmL/molK*298K / 1.017atm

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b. PV = nRT

P = nRT/V

P = 0.0947mol*0.082atmL/molK*309K/0.635L

P = 0.9216atm * (760mmHg/1atm) = 700mmHg

c. PV = nRT

PV/nR = T

P = 727mmHg * (1atm / 760mmHg) = 0.9566atm

0.9566atm*13.3L/0.393mol*0.082atmL/molK = T

T = 395K

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