Answer:
2.222 that is the answer i think might want to ask
O2 is the limited reactant
Answer:
19.264×
atoms are present in 3.2 moles of carbon.
Explanation:
It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.
So, as 1 mole of any element = Avagadro's number of atoms = 6.02×
atoms
It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.
As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.
1 mole of carbon = 6.02 ×
atoms
Then, 3.20 moles of carbon = 3.20 × 6.02 ×
atoms
Thus, 19.264×
atoms are present in 3.2 moles of carbon.
Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
![\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3DRT_b%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29-3.958%2B1.555lnP_c%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
Where,
is the Heat of vaoprization (J/mol)
is the normal boiling point of the gas (K)
is the Critical temperature of the gas (K)
is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:



Applying the above equation to find heat of vaporization as:
![\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3D8.314%5Ctimes307.4%20%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29-3.958%2B1.555ln36.4%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)

The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,

<u>Option C is correct</u>