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timurjin [86]
3 years ago
11

calculate the translational partition function of xenon and helium at the them temperature and volume

Chemistry
1 answer:
son4ous [18]3 years ago
4 0

Explanation:

a) The ratio of the translational partition of D2 and H2 is 2.82

b) The ratio of the translational partition of xenon and helium is 187.56

Explanation:

a) The translational partition for D₂ is equal to:

Where

m = mass of molecule

k = Boltzmann constant

T = absolute temperature

V = volume of container

h = Planck constant

If rewrite the equation:

The same for the H₂ molecule:

Taking the ratio D₂ and H₂:

Where

mH₂ = 2.015894 amu

mD₂ = 4.028204 amu

b) The same way to part a)

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Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas?
Pachacha [2.7K]
Na is the answer hope it help
7 0
3 years ago
How many moles are in 87.62 grams of strontium?
Zielflug [23.3K]

Explanation:

it's one mole because the atomic mass is treated as molar mass and the atomic mass of strontium is 87.62 grams/mol

5 0
3 years ago
A mixture of methane and krypton gases is maintained in a 8.99 L flask at a pressure of 1.68 atm and a temperature of 46 °C. If
sweet-ann [11.9K]

Answer:

22.81 g

Explanation:

Given that:

Pressure = 1.68 atm

Temperature = 46 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (46 + 273.15) K = 319.15 K

Volume = 8.99 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1.68 atm × 8.99 L = n × 0.0821 L.atm/K.mol × 319.15 K

<u>⇒n = 0.5764 moles </u>

Given that :

Amount of methane = 4.88 g  

Molar mass = 16.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{4.88\ g}{16.04\ g/mol}

Moles= 0.3042\ mol

<u>Moles of Krypton = Total moles - Moles of methane = 0.5764 - 0.3042 moles = 0.2722 moles</u>

Also, Molar mass of krypton = 83.798 g/mol

So,

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.2722\ moles=\frac{Mass}{83.798\ g/mol}

<u>Mass of krypton = 22.81 g</u>

6 0
4 years ago
(You do) If you have 47.2 mol of Na available, along with an excess of Cl₂, how many grams of NaCl can you produce?
IrinaVladis [17]

Answer:

2,760 grams NaCl

Explanation:

To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.

2 Na + Cl₂ --> 2 NaCl

Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol

Molar Mass (NaCl) = 58.44 g/mol

47.2 moles Na           2 moles NaCl              58.44 grams

----------------------  x  ---------------------------  x  -------------------------  =
                                   2 moles Na                   1 mole NaCl

= 2,758.368 grams NaCl

= 2,760 grams NaCl

5 0
2 years ago
Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co
amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as p_{d} and the proportion of air that is by-passed as p_{bp}, being p_{d}+p_{bp}=1.

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

0.004 = 0.016*p_{bp}+ 0.002*p_{d}

Replacing the first equation in the second one we have

0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0
4 years ago
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