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Andrej [43]
3 years ago
10

The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v).

Chemistry
2 answers:
vovangra [49]3 years ago
8 0

Answer:

Density is an intensive property

Explanation:

The intensive properties don´t depend on mass or size, if you have a liter of water, this litter have a density equal 1000 kg/m^3. If zo have 250 liters of water, the density is 1000 kg/m^3 too.

The density does not change with the addition or substraction of matter.

I hope I have been clear.

Bas_tet [7]3 years ago
5 0

Answer with explanation:

Extensive properties depend on the total matter while intensive properties do not, mass and volume are extensive properties because a bigger mass and a bigger volume are indicative of a more massive object, on the other hand, a more colorful or denser object is not an indication of a more massive object. Density, in particular, is a ratio so it will always be the same regardless of the size of the object, therefore, it is not directly related with the total amount of mass; to better understand this concept imagine you have a bag of sand and you add more sand to the bag, you are increasing the mass and the volume but because both increases the ration m/V do not.

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I need help with 1,2,3, and 4
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  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

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          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

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         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

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Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

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b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

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<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

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Divide both sides by 760 mmHg

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<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

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