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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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Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

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\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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