Answer:
Object 2 with a density of 0.68
Explanation:
object 2 has a lower density than water so it will float
Answer:
C. Carbon dioxide
Explanation:
Carbon dioxide is one of the end-product of combustion reactions involving many fuels today.
With the rapid increase in urbanization and technological development, man demand for energy increased tremendously. The discovery of fossil fuels paved the way for the astronomical increase in the concentration of carbon dioxide in the atmosphere. The burning of fossil fuels like coal and oil invovles the process where the carbon atoms present in these fuels combine with oxygen in the air to make CO2. This has resulted in an increase in the concentration of atmospheric carbon dioxide (CO2).
The burning fossil fuels for electricity, industry, heat, and transportation are the major sources of the emossion of carbon dioxide.
Also, the cutting down of trees for paper production, building construction and for the establishment of settlements also increase the concentration of carbon dioxide in the atmosphere. Trees are help remove carbon dioxide from the atmosphere through the process of photosynthesis. However, when these trees are cut down, carbon dioxide accumulates in the atmosphere.
Answer:
1.
was the
value calculated by the student.
2.
was the
of ethylamine value calculated by the student.
Explanation:
1.
The
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
value calculated by the student.
2.
The
value of ethylamine = 11.87


![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
of ethylamine value calculated by the student.
First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>