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Naily [24]
3 years ago
13

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr

essure conditions?
Boyle’s law
Charles’s law
Dalton’s law
Gay-Lussac’s law
Chemistry
2 answers:
lions [1.4K]3 years ago
8 0

Charles's Law i think

MAXImum [283]3 years ago
5 0

Answer is: Charles’s law.

Charle's law (the temperature-volume law): the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature:  V₁/T₁ = V₂/T₂.  

For example:

V₁(gas) = 20.0 L; initial volume.

T₁(gas) = 280 K.; initial temperature

V₂(gas) = 10.0 L; final volume

T₂(gas) = ?; final temperature.

20.0 L/ 280 K = 10.0 L / T₂.  

T₂ = 140 K.  

As the volume goes down, the temperature also goes down, and vice-versa.  

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6 0
3 years ago
What is the density of an object with 525 grams and a volume of 15cm3?
Hunter-Best [27]

Answer: 35 g/cm

Explanation:

Density equals mass over volume. 525 divided by 15 is 35

7 0
3 years ago
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Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric
kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas  is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

pressure\propto \frac{1}{volume} (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure  is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0
3 years ago
CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the
nirvana33 [79]

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

3 0
3 years ago
Which of these are examples of passive transport? A. osmosis B.diffusion c. endocytosis d.8 mitosis​
AfilCa [17]

Answer:

Passive Transport

Explanation:

The three examples of passive transport are

Diffuison

Osmosis

facilated diffuison

So the answer can be A or B

4 0
3 years ago
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