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satela [25.4K]
3 years ago
14

The density of benzene at 15 ∘C is 0.8787 g/mL. Calculate the mass of 0.1200 L of benzene at this temperature.

Chemistry
2 answers:
BabaBlast [244]3 years ago
8 0

The mass of 0.1200 L of benzene at 15° Celsius is 105.44 g.

FURTHER EXPLANATION

Density is a measure of how tightly packed the particles of a material are. The more tightly packed the particles, the greater the density and vice versa.  Density is also mostly defined as the amount of matter in a given volume or space. The amount of matter corresponds to its mass. Density is an intensive property of matter so it may be used to qualitatively identify an unknown substance.

To determine the mass of benzene, recall the definition of density:

density = \frac{mass}{volume}

Since the density and the volume of benzene are given, the mass can be calculated as follows:

mass = density \times volume

Before the given volume and density can be plugged into the equation, volume of benzene must first be converted to milliliters.

0.1200 \ L \times \frac{1000 \ mL}{1 \ L} = \boxed {120 \ mL}

Plugging into the equation for mass:

mass = 0.8787 \frac{g}{mL} \times 120 \ mL\\\\\boxed {mass = 105.444 \ g}

LEARN MORE

  • Learn more about Dimensional Analysis brainly.com/question/1557970
  • Learn more about Mole Conversion brainly.com/question/5297200
  • Learn more about Intensive property brainly.com/question/430934

<em>Keywords:  density, unit conversion, benzene</em>

Pavel [41]3 years ago
3 0
Use the density to convert volume into mass. 

since the density is in g/ml and the volume was given in Liters, we need to first convert the Liters into mililiters. just multiply by 1000 or move the decimal three times.

0.1200 Liters= 120.0 mL

120.0 mL (0.8787 grams/ 1 mL)= 105 grams
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The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

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$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

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