The factors such as structure, pressure and temperature on solubility has to be discussed. Henry law and the reason why Henry law does not work for Hydrochloric acid has to be explained. The terms hydrophobic and hydrophilic has to be explained. Concept Introduction: Henry’s law can be given as the quantity of gas dissolve in a solution is proportional to the pressure of the gas above the solution. Henry’s law gives the relationship between the pressure of the gas and the concentration of gas dissolved. The equation can be given as, C=kP where, C= Concentration of the gas dissolved k=constant P=Partial pressure of the solute in gaseous solute
The factors such as structure, pressure and temperature on solubility has to be discussed. Henry law and the reason why Henry law does not work for Hydrochloric acid has to be explained. The terms hydrophobic and hydrophilic has to be explained. Concept Introduction: Henry’s law can be given as the quantity of gas dissolve in a solution is proportional to the pressure of the gas above the solution. Henry’s law gives the relationship between the pressure of the gas and the concentration of gas dissolved. The equation can be given as, C=kP where, C= Concentration of the gas dissolved k=constant P=Partial pressure of the solute in gaseous solute
Solution Summary: The author explains Henry's law and the reason why it doesn't work for Hydrochloric acid.
Interpretation: The factors such as structure, pressure and temperature on solubility has to be discussed. Henry law and the reason why Henry law does not work for Hydrochloric acid has to be explained. The terms hydrophobic and hydrophilic has to be explained.
Concept Introduction:
Henry’s law can be given as the quantity of gas dissolve in a solution is proportional to the pressure of the gas above the solution.
Henry’s law gives the relationship between the pressure of the gas and the concentration of gas dissolved. The equation can be given as,
C=kP
where, C= Concentration of the gas dissolved
k=constant
P=Partial pressure of the solute in gaseous solute
1) Suppose 0.1 kg ice at 0°C (273K) is in 0.5kg water at 20°C (293K). What is the change in entropy of the ice as it melts at 0°?
To produce the original "water gas" mixture, carbon (in a combustible form known as coke) is reacted with steam: 131.4 kJ + H20(g) + C(s) → CO(g) + H2(g) From this information and the equations in the previous problem, calculate the enthalpy for the combustion or carbon to form carbon dioxide.
kindly show me how to solve both parts of the same long problem. Thanks
we were assigned to dilute 900ppm
in to 18ppm by using only 250ml vol
flask. firstly we did calc and convert
900ppm to 0.9 ppm to dilute in 1 liter.
to begin the experiment we took
0,225g of kmno4 and dissolved in to
250 vol flask. then further we took 10
ml sample sol and dissolved in to 100
ml vol flask and put it in to a
spectrometer and got value of 0.145A
.
upon further calc we got v2 as 50ml
. need to find DF, % error (expval and
accptVal), molarity, molality. please
write the whole report. thank you
The format, tables, introduction,
procedure and observation, result,
calculations, discussion and
conclusion
Q5. Predict the organic product(s) for the following transformations. If no reaction will take place
(or the reaction is not synthetically useful), write "N.R.". Determine what type of transition state
is present for each reaction (think Hammond Postulate).
I
Br₂
CH3
F2, light
CH3
Heat
CH3
F₂
Heat
Br2, light
12, light
CH3
Cl2, light
No
Chapter 11 Solutions
Student Solutions Manual for Zumdahl/Zumdahl/DeCoste?s Chemistry, 10th Edition