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Drupady [299]
2 years ago
11

There is an extreme temperature difference between day and night in a desert during clear weather. However, even though the outs

ide temperature drops greatly at night, the water in a swimming pool in the desert tends to remain warm. Use your knowledge of heat energy, specific heat, and heat transfer to explain these facts.
please help!!!!!! ​
Chemistry
1 answer:
den301095 [7]2 years ago
5 0

Sand has a lower heat capacity hence it heats up easily and looses heat easily. On the other hand, water has a higher heat capacity and looses heat more slowly hence it remains warm at night.

<h3>What is heat transfer?</h3>

The transfer of heat refers to the movement of heat from one body to another. We should note that sand has a low specific heat capacity hence it is rapidly heated up in the day and looses heat quickly at night.

On the other hand, water has a high specific heat capacity hence it heat up slowly and also does not loose heat easily even when the surrounding temperature is much colder. As such, there is an extreme temperature difference between day and night in a desert during clear weather and the water in a swimming pool in the desert tends to remain warm even when the outside temperature drops greatly at night.

Learn more about heat transfer: brainly.com/question/12107378

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potassium chlorate (kclo3) decomposes into potassium chloride (kcl) and oxygen gas (o2) how many grams of oxygen can be produced
sergeinik [125]

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m_{O_2}=354.24gO_2

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)

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m_{O_2}=7.38molKClO_3*\frac{3molO_2}{2molKClO_3}*\frac{32gO_2}{1molO_2} \\\\m_{O_2}=354.24gO_2

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The colligative molality of an unknown aqueous solution is 1.56 m.
yawa3891 [41]

Answer:

Vapor pressure of solution = 17.02 Torr

T° of boiling point for the solution is 100.79°C

T° of freezing point for the solution is -2.9°C

Explanation:

Let's state the colligative properties with their formulas

- <u>Vapor pressure lowering</u>

ΔP = P° . Xm . i

- <u>Boiling point elevation</u>

ΔT = Kb . m . i

-<u> Freezing point depressión</u>

ΔT = Kf . m . i

ΔP = Vapor pressure pure solvent (P°) - Vapor pressure solution

ΔT = T° boling solution - T° boiling pure solvent

ΔT = T° freezing pure solvent - T° freezing solution

i represents the Van't Hoff factor (ions dissolved in the solution). If we assume that the solute is non-volatile and the solution is ideal i = 1

Kf and Kb are cryoscopic and ebulloscopic constant, they are  specific to each solvent.

Vapor pressure works with mole fraction (Xm) and the only data we have is molality, so we consider 1.56 moles of solute and 1000 g of solvent mass.

Moles of solvent → solvent mass / molar mass of solvent

Moles of solvent → 1000 g / 18 g/mol = 55.5 moles

Mole fraction is moles of solute / Total moles (mol st + mol sv)

Mole fraction: 1.56 / (1.56 + 55.5) = 0.027

- Vapor pressure lowering

ΔP = P° . Xm . i

17.5 Torr - Vapor pressure of solution = 17.5 Torr . 0.027 . 1

Vapor pressure of solution = - (17.5 Torr . 0.027 . 1 - 17.5 Torr)

Vapor pressure of solution = 17.02 Torr

- Boiling point elevation

ΔT = Kb . m . i

T° boiling solution - 100° = 0.512 °C/ m . 1.56 m . 1

T°boiling solution = 0.512 °C/ m . 1.56 m . 1 + 100°C

T°boiling solution = 100.79°C

- Freezing point depression

ΔT = Kf . m . i

0°C - T° freezing solution = 1.86 °C/m . 1.56 m . 1

T° freezing solution = - (1.86 °C/m . 1.56 m)

T° freezing solution = -2.9°C

3 0
3 years ago
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