The boiling point and freezing point has to be calculated. Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties. The elevation in boiling point can be given by the equation, ΔT=K b m solute Where, ΔT = change in boiling point elevation K b = molal boiling point elevation constant m solute = molality of solute The depression in freezing point can be given by the equation, ΔT=K f m solute Where, ΔT =change in freezing point depression K f = molal freezing point depression constant m solute = molality of solute
The boiling point and freezing point has to be calculated. Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties. The elevation in boiling point can be given by the equation, ΔT=K b m solute Where, ΔT = change in boiling point elevation K b = molal boiling point elevation constant m solute = molality of solute The depression in freezing point can be given by the equation, ΔT=K f m solute Where, ΔT =change in freezing point depression K f = molal freezing point depression constant m solute = molality of solute
Solution Summary: The author explains that the boiling point and the freezing point are together known as colligative properties. The molarity of ionized Formic acid solution is calculated by the equation.
Interpretation: The boiling point and freezing point has to be calculated.
Concept Introduction: The depression in freezing point, the elevation of boiling point and osmotic pressure are together known as colligative properties.
The elevation in boiling point can be given by the equation,
ΔT=Kbmsolute
Where,
ΔT= change in boiling point elevation
Kb = molal boiling point elevation constant
msolute = molality of solute
The depression in freezing point can be given by the equation,
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3)
→
Give the expression for the acceptable rate.
→
→
(A).
d[N205]
dt
==
2k,k₂[N₂O₂]
k₁+k₁₂
(B).
d[N2O5]
=-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³
dt
(C).
d[N2O5]
=-k₁[N₂O] + k [NO] - k₂[NO] [NO]
d[N2O5]
(D).
=
dt
= -k₁[N2O5] - k¸[NO][N₂05]
dt
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
R lactam or lactone considering as weak acid or weak base and why
81. a. Propose a mechanism for the following reaction:
OH
CH2=CHCHC=N
b. What is the product of the following reaction?
HO
H₂O
N=CCH2CH2CH
OH
HO
CH3CCH=CH2
H₂O
C=N
82. Unlike a phosphonium ylide that reacts with an aldehyde or a ketone to form an alkene a sulfonium ulia