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Norma-Jean [14]
2 years ago
15

Calculate, by explicit summation, the vibrational partition functionand the vibrational contribution to the molar internal energ

y of I2 Molecule At (a) 100 K, (b) 298 K given that its vibrational energy levels lie at thefollowing wavenumbers above the zero-point energy level: 0, 213.30, 425.39,636.27, 845.93 cm−1. What proportion of I2molecules are in the ground andfirst two excited levels at the two temperatures

Chemistry
1 answer:
Sphinxa [80]2 years ago
3 0

Answer:

Explanation:

Find attached the solution

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Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
2 years ago
When metals combine with nonmetals, the metallic atoms tend to 1. lose electrons and become positive ions 2. lose electrons and
Vika [28.1K]
The correct answer is 1. Lose electrons and become positive ions.


I hope my answer was beneficial to you! c:
5 0
2 years ago
Read 2 more answers
What is the enthalpy for reaction 1 reversed?reaction 1 reversed: N2O4→N2 + 2O2
Nutka1998 [239]

Answer:

8kJ/mol

Explanation:

since the forward reaction is -8kJ/mol, the backward reaction has the same enthalphy but reversed

7 0
1 year ago
How many liters of cane juice is needed to supply 5g sucrose if cane juice contains 12% sucrose
pickupchik [31]

Given parameters:

Mass of sucrose  = 5g

Density of sucrose  = 1.12g/mL

Percentage of sucrose per liter of cane juice  = 12%

Unknown:

Volume of cane juice needed = ?

We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.

Mathematically;

                Density  = \frac{mass}{volume}

Now solve for the volume of sucrose;

                  1.12g/mL = \frac{5}{Volume}

       Volume  = \frac{5}{1.12}    =  4.46mL   = 4.46 x 10⁻³L since 1000mL  = 1L

Since 12% of 1 liter of cane juice is sucrose;

                    12% of  x  liter of cane juice  = 4.46 x 10⁻³L

                   Volume of cane juice  = 4.46 x 10⁻³ x \frac{100}{12}   = 0.037L

Volume of cane juice is  0.037L

   

6 0
2 years ago
As you may know, ethyl alcohol, C2H5OH, can be produced by the fermentation of grains, which contain glucose, C6H12O6 → +2C2H5OH
fredd [130]

Answer:

a. 510.6 g of C₂H₅OH are produced from 1kg of glucose

b. 171.1 g of glucose are required

Explanation:

Chemist reaction is this:

C₆H₁₂O₆  →  2C₂H₅OH(l) + 2CO₂(g)

So 1 mol of glucose can produce 1 mol of ethyl alcohol.

First of all, we should convert the mass to g, afterwards to moles

1 kg . 1000 g/ 1kg = 1000 g . 1 mol/180 g = 5.55 moles

Then we can think, this rule of three

1 mol of glucose can produce 2 moles of ethyl alcohol

Then 5.55 moles of glucose may produce the double of moles of C₂H₅OH

(5.55 .2)/1 = 11.1 moles.

Let's convert the moles to mass → 11.1 mol . 46g /1mol = 510.6 g

b. Let's determine the liters of ethyl alcohol we need.

1 gasohol is 10 mL C₂H₅OH / 90 ml of gasoline. We should make a rule of three.

In 90 mL of gasoline we have 10 mL of C₂H₅OH

In 1000 mL (1L) we would have (1000 . 10)/ 90 = 111.1 mL

Now we have to determine the mass of C₂H₅OH that is contained in the volume we have calculated. We must use the density.

Density = Mass /Volume

0.79 g/mL = Mass / 111.1 mL

0.79 g/mL . 111.1 mL = 87.7 g

Now, we convert the mass to moles → 87.7 g . 1mol/ 46g = 1.91 mol

Ratio is 2:1 so 2 moles of C₂H₅OH are produced by 1 mol of glucose

Therefore 1.91 mol would be produced by (1.91 .1)/2 = 0.954 moles

Finally we convert the moles of glucose to mass:

0.954 mol . 180 g/ 1mol = 171.7 grams.

5 0
2 years ago
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