ln (760mmHg / x) = (40.7KJ/mol) / 8.314J/K*mol) * (1/358K - 1/373K) Solve for x
ln (760mmHg / x) = (40.7KJ/mol) / 8.314J/K*mol) * (1/358K - 1/373K)
ln (760mmHg / x) = (40700J/mol) / 8.314J/K*mol) * (0.00279K - 0.00268K)
ln (760mmHg / x) = (40700J/mol) / 8.314J/K*mol) * (0.00279K - 0.00268K)
= 40700J/mole*(0.00279K - 0.00268K) /8.314J/K*mole
= 40700J/mole*0.00011K/8.314J/K*mole
= 0.5385
ln (760mmHg / x) = 0.5385
(760mmHg / x) = e^0.5385
(760mmHg / x) = 1.7134
x = 760mmHg/1.7134
x = 443.56mmHg>>>>answer
ln (760mmHg / x) = (40.7KJ/mol) / 8.314J/K*mol) * (1/358K - 1/373K) Solve for x
Ln(760mmHg/x) = (40.7KJ/mol) / (8.314J/K*mol) * (1/358K - 1/373K) solve for x
11. Calculate the percentage of each conformer at 23 °C (AEE-RTINK, where R= 8.314J/K mol), ACH₃ н, Сн HC НАС HC CH3 Conformer A Conformer B Interaction Energy Cost (kJ/mol H-H eclipse 4.0 H-CH3 eclipse CH3-CH: eclipse 11.0 CH3-CH3 gauche 3.8 6.0
.5LN 2= -LN | CSCX-COTX | +( LN ((SQUARE ROOT 2) +1|)). solve for X
If g(x)=−ln(1−x) , what is g^(k)(0) (for k=1,2,3,… ) ?options:a) 1/kb)1/1-kc)k!d)(k-1)!e) (k+1)!
Solve the following using Optimization 1) Using substitution: U = 2 ln(x) + ln(y) s.t. 2x + y = 20 2) Using LaGrangian: U = √x + ln(y) s.t. x+y=10 3) Using both: U = 5 ln(x) + 3 ln(y) s.t. 4x + 3y = 100
8.314J. 1 atm- 1.01325X10% Pa; 1 m2-101:760 torr-1 atm; R-0.0821 1. atm. K' mol K.1 mol Note that the different parts of question 1 are independent. Answering a part of a question wrong or skipping it altogether should not affect your ability to answer other parts correctly. iven the relationship dG-V dP-S dT which indicates that G-(P, T). Also knowing that G is a state functions (meaning the reciprocity rule applies on it), derive the relation: os OP av Using...
The vapor pressure of Y is 4.050 atm at 290K and 13.365 atm at 373K. What are the standard free energies of vaporization at the two temperatures? Hint: vaporization reaction is Y(I) = Y(g). «J/mol AG290º = AG373° = kJ/mol Assume that the standard entropy and enthalpy of vaporization of Y are independent of temperature and determine their values J/(mol-K) AS° = AH° = kJ/mol What is the normal boiling point of Y in degrees Celcius? 1 t= t =...
For the reaction: C → D +B the graph of ln k vs 1/T(K) gives the equation: y = -8.1x103x + 45.0. Using this information, calculate the activation energy, (J/mol)
-k=ln(1/3) so k=ln(3) how come it makes sense? what kind of ln Rule does it use?
8. With K(t) = ln(EetX), show that k'(0)-EX], K"(0) = Var(X).