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lukranit [14]
2 years 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]2 years 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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Identify the factors that directly favor the unloading of oxygen from hemoglobin in the blood near metabolically active tissues.
AlexFokin [52]

Answer:

A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES

B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.

C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.

Explanation:

Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.

There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.

The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.

There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.

The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.

7 0
2 years ago
From the Bohr equation in the introduction, the calculated energy of an electron in the sixth Bohr orbit of a hydrogen atom is
Natalka [10]

Answer:

<em><u>= - 0.38 eV</u></em>

Explanation:

Using Bohr's equation for the energy of an electron in the nth orbital,

E = -13.6 \frac{Z^{2} }{n^{2} }

Where E = energy level in electron volt (eV)

Z = atomic number of atom

n = principal state

Given that n = 6

⇒ E = -13.6 × \frac{1^{2} }{6^{2} }

<em><u>= - 0.38 eV</u></em>

<em><u></u></em>

<em>Hope this was helpful.</em>

<em><u></u></em>

4 0
2 years ago
A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)
allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0
2 years ago
Read 2 more answers
Compare the composition of sucrose purified from the sugar cane with the composition of sucrose purified from sugar beets.
Vladimir79 [104]

Answer: they are the same.


Explanation:


1) Sucrose is a compound with chemical formula C₁₂H₂₂O₁₁


2) That means that all molcules of sucrose will have the same kind of atoms in the same proportion, whic is to say same composition:


12 atoms of C: 22 atoms of H: 11 atoms of O, per each molecule of sucrose.


3) For this question you can rely in the definition of compound: a pure substance formed by the combination of two or more elements always in the same ratio (same composition).


4) That also implies, that all the molecules of sucrose have the same properties.

3 0
2 years ago
Read 2 more answers
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
2 years ago
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