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Solnce55 [7]
2 years ago
7

Catie set a glass of hot water and a glass of cold water on the kitchen table and let them sit for one hour. How will the partic

le motion within the glasses change?A.The particles in both glasses will speed up.B.The particles in both glasses will slow down.C.The particles in the hot water will slow down, while the particles in the cold water will speed up.D.The particles in the hot water will speed up, while the particles in the cold water will slow down.
Chemistry
1 answer:
Elis [28]2 years ago
5 0

Answer:

C.The particles in the hot water will slow down, while the particles in the cold water will speed up

Explanation:

Heat causes changes in the molecules of substances when applied to them. The molecules of water, as a case study used in this question, when heated has an increased kinetic energy which causes its particles to move faster i.e. speed up. Also, when water is freezed or cold, it's particles move slower i.e. slow down due to decreased kinetic energy.

Hence, when a hot glass of water and a cold glass of water are placed on a kitchen table for 1 hour, they don't remain in their respective thermal state.

- The hot water (with an increased movement of molecules) begin to cool down, hence, its particles slows down as it gets cold or less hot.

- Likewise, the cold water (with a decreased movement of molecules) begins to get warmer, hence, it particles speedens up as it gets warmer.

Therefore, The particles in the hot water will slow down, while the particles in the cold water will speed up

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Small children are occasionally injured when they try to inhale helium from
sattari [20]

Answer:

260 moles of Helium

Explanation:

V = 50L

T = 20°C = (20 + 273.15)K = 293.15K

P = 125 atm

R = 0.082 L.atm / mol. K

n = ?

From ideal gas equation,

PV = nRT

P = pressure of a given gas

V = volume it occupies

n = number of moles

R = ideal gas constant

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PV = nRT

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n = (125 * 50) / (0.082 * 293.15)

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8 0
3 years ago
4. A 14.5-L balloon is in the air where it is 20.0 C at 0.980-atm. A wind passes over and the balloon's pressure becomes 740.0-m
jeka94

The temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

<h3>What is Combined gas law?</h3>

Combined gas law put together both Boyle's Law, Charles's Law, and Gay-Lussac's Law. It states that "the ratio of the product of volume and pressure and the absolute temperature of a gas is equal to a constant.

It is expressed as;

P₁V₁/T₁ = P₂V₂/T₂

Given the data in the question;

  • Initial volume V₁ = 14.5L
  • Initial pressure P₁ = 0.980atm
  • Initial temperature T₁ = 20.0°C = 293.15K
  • Final volume V₂ = 14.3L
  • Final pressure P₂ = 740.mmHg = 0.973684atm
  • Final temperature T₂ = ?

We substitute our given values into the expression above.

P₁V₁/T₁ = P₂V₂/T₂

( 0.980atm × 14.5L )/293.15K = ( 0.973684atm × 14.3L )/T₂

14.21Latm / 293.15K = 13.92368Latm / T₂

14.21Latm × T₂ = 13.92368Latm × 293.15K

14.21Latm × T₂ = 4081.72679LatmK

T₂ = 4081.72679LatmK / 14.21Latm

T₂ = 287.24K

T₂ = 14.09°C

Therefore, the temperature of the wind as that decreases the volume and the pressure of the balloon to the given values is 14.09°C.

Learn more about the combined gas law here: brainly.com/question/25944795

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Who said atoms are neither created nor destroyed in chemical reactions?
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