Answer:
i

ii

Explanation:
From the question we are told that
The first temperatures is 
The second temperature is 
Generally the equation for the most highly populated rotational energy level is mathematically represented as
![J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}](https://tex.z-dn.net/?f=J_%7Bm%7D%20%3D%20%5B%20%5Cfrac%7BRT%7D%7B2B%7D%5D%20%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B2%7D)
Here R is the gas constant with value 
Also
B is given as 
Generally the energy require per mole to move 1 cm is 12 J /mole
So
will require x J/mole

=> 
So at the first temperature
![J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5](https://tex.z-dn.net/?f=J_%7Bm%7D%20%3D%20%5B%20%5Cfrac%7B8.314%20%2A%20298%20%20%7D%7B2%2A%20%202.928%20%7D%5D%20%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20-%200.5%20)
=> 
So at the second temperature
![J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5](https://tex.z-dn.net/?f=J_%7Bm%7D%20%3D%20%5B%20%5Cfrac%7B8.314%20%2A%20373%20%20%7D%7B2%2A%20%202.928%20%7D%5D%20%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20-%200.5%20)
=> 
Answer:
163.94
Explanation:
Na = 22.99 (3)
P = 30.97
O = 16.00 (4)
22.99(3) + 30.97 +16.00(4) = 163.94
Hope that helps
Answer : The enthalpy change for the reaction is, 419.5 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The given chemical reaction is,

Now we have to determine the enthalpy change for the reaction below:

By reversing and then dividing the reaction by 2, we get the enthalpy change for the reaction.
The expression will be:



Therefore, the enthalpy change for the reaction is, 419.5 kJ
Answer: 21.5kg in grams is 21,500 grams
21.5kg in mg is 215,000,00
Explanation:
Answer:
The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.
Explanation:
Number of hours worker exposed to xylene = 
The concentration of xylene in the workplace =
The worker is inhaling air at a rate of
.
Amount xylene inhaled by worker in an hour :
= 
Amount xylene inhaled by worker in 320 hours:

1 μg = 0.001 mg
Amount xylene inhaled by worker in 320 hours = 11.520 mg
1 day = 24 hours
Amount xylene inhaled by worker in 1 day:

Assuming 70 kg body mass, the chronic daily intake of xylene :

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.