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>
Answer:
The answer to your question is D. 25 grams.
<span>Fe(NO3)2
The NO3 part is a poly-atomic ion with total charge -1.
This is because Fe has a +2 charge and two NO3's with a -1 charge will balance out to 0.
Most often we just make the assumption that Oxygen has a -2 oxidation number because it is very electro-negative.
So to find N, we just need an oxidation number that balances out with 3(-2) to get -1 (the total charge of the ion)</span>
Answer:
The cylindrical tube can hold 0.215 cubic decimeters of blood.
Explanation:
Diameter of the cylindrical tube = d = 4.5 cm
r = d/2 ,r = 2.25 cm
Height of the cylindrical tube = h = 13.5 cm
Volume of the cylinder = V = 
The cylindrical tube can hold 0.215 cubic decimeters of blood.
Answer:
Redox reaction and single displacement
Explanation:
This reaction is first of all a redox reaction. A redox reaction is a reaction that involves both oxidation and reduction. Oxidation involves increase in oxidation number while reduction involves decrease in oxidation number.
Copper (Cu) had an oxidation number of "0" as a reactant but had an oxidation number of "2+" in the product [Cu(NO₃)₂] hence oxidation occurred.
Nitrogen (N) had an oxidation number of "5+" in the reactant (HNO₃) but had an oxidation number of "4+" in the product (NO₂) hence reduction also occurred.
Also, from the reaction, it can be deduced that copper (Cu) displaced hydrogen (H) from the nitric acid (HNO₃) solution to form copper (II) nitrate [Cu(NO₃)₂]. It should be noted that copper can displace hydrogen because it is higher than hydrogen in the electrochemical series. Hence, this reaction can also be called a single displacement reaction. A single displacement reaction is a reaction in which an atom of an element replaces another atom in a compound (as seen in the equation given in the question).
