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lukranit [14]
1 year ago
5

Carbon dioxide, CO₂, changes from the solid phase to the gas phase at 1 atm and 194.5 K. In the solid phase, CO₂ is often called

dry ice. When dry ice sublimes in air at 298 K, the water vapor in the air can condense, forming a fog of small water droplets. This fog is often used for special effects at concerts and in movie-making.
1. State the direction of heat flow between the dry ice and the water vapor in the air.
2. At 1 atm and 298 K, compare the potential energies of the water molecules before and after the water vapor condenses.
3. At 1 atm and 190. K, compare the amount of thermal energy in a 1.0-kilogram block of dry ice to the amount of thermal energy in a 2.0-kilogram block of dry ice.
Chemistry
1 answer:
FinnZ [79.3K]1 year ago
5 0

Answer:

1. From water vapor to the dry ice;

2. The potential energy is higher before the water vapor condenses;

3. The thermal energy is higher in the 2.0 kg block.

Explanation:

1. The heat flows from the system with high temperature to the system with low temperature. The water vapor is at 298 K, and the dry ice is at 194.5 K.

2. The energy of the molecules is related to the temperature and the physics state. At the gas state, the molecules are more agitated, and the energy is higher than the liquid state. So, when the vapor condenses to a liquid, the energy decreases.

3. The thermal energy can be calculated by:

Q = m*c*ΔT

Where m is the mass, c is the specific heat, and ΔT the variation in the temperature. So, when the mass increase, thermal energy also increases.

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1 year ago
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A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is
musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁  = 1.08atm

V₁  = 250mL

T₁  = 24°C

P₂  = 2.25atm

T₂  = 37.2°C

V₂  = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

              \frac{P_{1} V_{1} }{T_{1} }   =  \frac{P_{2} V_{2} }{T_{2} }

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

        T₁  = 24°C,  T₁   = 24 + 273 = 297K

        T₂  = 37.2°C , T₂  = 37.2 + 273  = 310.2K

Input the variables and solve for V₂

        \frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}

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6 0
2 years ago
Explain how the models you developed show that when methane combines with oxygen to form carbon dioxide and water, no atoms are
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Answer: the bonds in the methane and oxygen come apart, the atoms rearrange and then re-bond to form water and carbon dioxide

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1 year ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
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Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

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