Answer:
E
Explanation:
Here in this question, what we will do is to select which of the pairs that do not correlate.
A. Enthalpy and heat content
This two terms are at par with each other. By definition, the enthalpy of a system simply is the total amount of heat content it has.
B. Endothermic reaction and +H
These two terms are at par with each other. An endothermic reaction is one in which heat is absorbed from the surroundings. It has a positive value for the heat content i.e the enthalpy is positive and thus H is positive.
C. Exothermic reaction and -H
An exothermic reaction is one in which heat is released to the environment. It usually has a negative value for the enthalpy and thus the value of H is negative.
D. High energy and High Stability
These two terms are not at par. When an entity has or is of high energy, it is usually unstable. An entity at a higher energy level will not be stable until it goes to a lower level of energy.
Thus higher energy level is associated with lesser stability while lower energy levels are associated with higher stability. The lesser the energy of an entity, the higher its stability.
This makes the option our answer.
Hey there!
H₃PO₄
Find molar mass.
H: 3 x 1.008 = 3.024
P: 1 x 30.97 = 30.97
O: 4 x 16 = 64
---------------------------------
97.994 grams
The mass of 1 mole of H₃PO₄ is 97.994 grams.
We have 4.5 moles.
97.994 x 4.5 = 440
The mass of 4.5 moles of H₃PO₄ is 440 grams.
Hope this helps!
The answer is 615.91 grams of <span>n2f4
Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
=615.91 grams
Explanation:
Relation between pressure, latent heat of fusion, and change in volume is as follows.
Also, 
where, 
is the difference in specific volumes.
Hence, 
As, 
= 22.0 J/mol K
And, 
...... (1)
where, 
= density of water
= density of ice
M = molar mass of water = 
Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.
= 
= 
Therefore, calculate the required pressure as follows.
= 
or, = 145 bar/K
Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = 
= 681.5 bar
Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.
