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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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Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after
Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

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Draw a diagram using H+ signs and OH- symbols of a solution that is acidic. Hint: Use the
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More positive ions shows acidic whereas more negative ions indicates basic solution.

<h3>Which charge show acidic solution?</h3>

That side which has more positive charges is considered as acidic solution while on the other hand, that region where negative charges are present in large number as compared to positive charges is considered as basic or alkaline solution.

So we can conclude that more positive ions shows acidic whereas more negative ions indicates basic solution.

Learn more about charge here: brainly.com/question/25923373

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