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Semenov [28]
2 years ago
13

A solution contains 3.95 g of carbon disulfide (CS2, molar mass = 76.13 g/mol) and 2.43 g of acetone ((CH3)2CO, molar mass = 58.

08 g/mol). The vapor pressure of pure carbon disulfide and acetone at 35 o C are 515 torr and 332 torr, respectively. Assuming ideal solution behavior, calculate the vapor pressure of each of the components and the total vapor pressure above the solution. The total vapor pressure above the solution at 35 o C was experimentally determined to be 645 torr. Is the solution ideal? If not, indicate whether the solution deviates from Raoult’s law in a positive or negative manner
Chemistry
1 answer:
Feliz [49]2 years ago
6 0

Answer:

The solution is not ideal and shows a positive deviation from Raoult’s law since Psolution (experimental) > Psolution (actual).

Explanation:

Number of moles of CS2 = 3.95g/76.13gmol-1 = 0.0519 moles

Number of moles of acetone = 2.43g/58.08gmol-1 = 0.0418 moles

Total number of moles = 0.0937 moles

Mole fraction of CS2 = 0.0519/0.0937 = 0.5538

Mole fraction of acetone = 0.0418/0.0937 = 0.4461

From Raoult’s law;

PCS2 = 0.5538 × 515 torr = 285.207 torr

Pacetone = 0.4461 × 332 torr = 148.1052 torr

Total pressure = 285.207 torr + 148.1052 torr = 433.3 torr

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what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.730 L}{301K}=\frac{V2}{273K}

Solving:

V2=273K*\frac{0.730L}{301K}

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

6 0
3 years ago
12. A beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. Will it float in water,
Greeley [361]

Taking into account the definition of density and Archimedes' principle, the beginner bowling ball will float on the water.

But first it is neccesary to know that density is a quantity referred to the amount of mass in a certain volume of a substance or a solid object.

In other words, the density is the relationship between the weight (mass) of a substance and the volume that the same substance occupies.

The expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

In this case, a beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. This is:

  • mass= 4.9 kg= 4900 g (being 1 kg= 1000 kg)
  • volume= 5.4 L= 5400 mL (being 1L=1000 mL)

Replacing in the definition of density:

density=\frac{4900 g}{5400 mL}

Solving:

<u><em>density=0.907 </em></u>\frac{g}{mL}<u><em /></u>

On the other hand, Archimedes' principle says that an object immersed in a liquid experiences an upward vertical force equal to the weight of the volume of the dislodged liquid.

The sinking or floating of an object is determined by its density with respect to that of the liquid in which it is submerged.

Considering water as the liquid where the object is submerged in this case, an object with a higher density than water will sink. In contrast, an object with a lower density than water will float.

In this case, considering that water has a density of 1 \frac{g}{mL}, the bowling ball for beginners has a lower density. This indicates that, having a lower density than water, the object will float.

In summary, the beginner bowling ball will float on the water.

Learn more about density:

  • <u>brainly.com/question/952755?referrer=searchResults </u>
  • <u>brainly.com/question/1462554?referrer=searchResults</u>

5 0
2 years ago
What is the electron configuration for<br> 08<br> 16
Elis [28]

The electron configuration for Oxygen : [He] 2s²2p⁴

<h3>Further explanation</h3>

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d, and f. The maximum number of electrons for each sub-shell is  

• s: 2 electrons  

• p: 6 electrons  

• d: 10 electrons and  

• f: 14 electrons  

Charging electrons in the sub-shell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

The element is Oxgen, with symbol O, and :

the atomic number=8=number of electron

the atomic mass=16

The electron configuration based on the number of electrons(for Oxygen=8), so the configuration :

\tt _8^{16}O:1s^22s^22p^4 or we can write with noble gas [He] 2s²2p⁴

3 0
2 years ago
How can the knowledge of a material’s chemical makeup or use be utilized?
shutvik [7]

To identify minerals

Each material chemical makeup are a variety of chemical compounds which has each own category. They have different functional groups which helps people identify which material they look for. For instance, hydroxyl group has chemical makeup of alcohol

6 0
3 years ago
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A dry mixture of KNO3 and sand could be separated by
Tatiana [17]
Place the mixture in hot water and stir well. 
<span>The KNO3 is very soluble in hot water. </span>
<span>Use a fine filter paper and filter off the sand. </span>
<span>The sand will be separated from the KNO3 solution. </span>
<span>The water can now be evaporated from the solution by further, gentle heating leaving the solid in the container.</span>
7 0
3 years ago
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