(e). The plot of ln(k) against 1/T will be a straight line.

(f). For first order in A, rate=k[A], so initial rate against [A] will be a straight line.

## Search found 56 matches

- Tue Mar 13, 2018 10:44 am
- Forum: General Rate Laws
- Topic: 15.99 (e)(f)
- Replies:
**3** - Views:
**219**

- Tue Mar 13, 2018 10:37 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Arrhenius Equation, Temperature, and Rate Constants
- Replies:
**1** - Views:
**179**

### Re: Arrhenius Equation, Temperature, and Rate Constants

Arrhenius equation: k=Ae^(-Ea/RT)

From it, you get the equation: ln(k2/k1)=(-Ea/R)(1/T2-1/T1)

If the problem asks you to solve k or Ea at one T, use the first equation. If the tells you to find k for a certain reaction at different reaction, use the second equation.

From it, you get the equation: ln(k2/k1)=(-Ea/R)(1/T2-1/T1)

If the problem asks you to solve k or Ea at one T, use the first equation. If the tells you to find k for a certain reaction at different reaction, use the second equation.

- Tue Mar 13, 2018 10:20 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.85
- Replies:
**1** - Views:
**137**

### Re: 15.85

A transitional state is an arrangement of molecules or atoms that can either go to form products or back to reactants. When you want to draw such structure, just connect all atoms or molecules with dashes (lines are used to connect atoms within a molecule).

- Tue Mar 06, 2018 10:46 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Four Cases
- Replies:
**2** - Views:
**186**

### Re: Four Cases

1. ΔS>0, ΔH>0, spontaneous at high T (TΔS>ΔH)

2. ΔS>0, ΔH<0, spontaneous at all T

3. ΔS<0, ΔH>0, non-spontaneous at all T (reverse rxn. is spontaneous)

4. ΔS<0, ΔH<0, spontaneous at low T

My approach is to remember the formula ΔG=ΔH-TΔS and see how can I get a negative ΔG.

2. ΔS>0, ΔH<0, spontaneous at all T

3. ΔS<0, ΔH>0, non-spontaneous at all T (reverse rxn. is spontaneous)

4. ΔS<0, ΔH<0, spontaneous at low T

My approach is to remember the formula ΔG=ΔH-TΔS and see how can I get a negative ΔG.

- Tue Mar 06, 2018 10:37 am
- Forum: General Rate Laws
- Topic: How the concentration of a reactant affects the rate of the reaction
- Replies:
**3** - Views:
**156**

### Re: How the concentration of a reactant affects the rate of the reaction

Based on the order of a reactant, the rate of the reaction can be affected by its concentration or its concentration squared (first and second order respectively). In case of zero order, its concentration would not affect the rate at all.

- Tue Mar 06, 2018 10:28 am
- Forum: Zero Order Reactions
- Topic: Slope From
- Replies:
**5** - Views:
**398**

### Re: Slope From

For zero order, y-intercept is initial concentration and slope is -k (rate constant) For first order, y-intercept is the natural logarithm of initial concentration and slope is -k For second order, y-intercept is the inverse of initial concentration and slope is k The slope-intercept form just help ...

- Tue Feb 27, 2018 11:06 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.19
- Replies:
**3** - Views:
**244**

### Re: 15.19

You do need to calculate the order for each reactant. To find the overall order of the reaction, you add up the order of each reactant.

In this case, A is first-order, B is second-order, and C is also second-order. So, the overall order is 5.

In this case, A is first-order, B is second-order, and C is also second-order. So, the overall order is 5.

- Tue Feb 27, 2018 11:02 am
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Midterm 6A (Multiple Choice)
- Replies:
**4** - Views:
**427**

### Re: Midterm 6A (Multiple Choice)

That is the entropy for gases are much larger than liquids and solids. Compare to this, the size of molecules has less effect.

- Tue Feb 27, 2018 10:57 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 14.47
- Replies:
**3** - Views:
**232**

### Re: 14.47

Yes, the actual number should be 3269017. The answer gives 10^6 because this part is the most significant one.

- Tue Feb 20, 2018 10:28 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.17 - HW
- Replies:
**2** - Views:
**191**

### Re: 14.17 - HW

You can find these half reactions in Appendix 2B.

- Tue Feb 20, 2018 10:24 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.17 HW problem
- Replies:
**1** - Views:
**89**

### Re: 14.17 HW problem

Because this redox reaction takes place in an acidic solution, so you add H+ to balance the equation.

- Tue Feb 20, 2018 10:21 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Question 23 part a
- Replies:
**1** - Views:
**165**

### Re: Question 23 part a

When you write out the half-reactions,

oxidation: 2Hg --> Hg2^2+ +2e-

reduction: NO3- + 4H+ + 3e- --> NO + 2H2O

The lowest possible number of moles of electrons is 6.

oxidation: 2Hg --> Hg2^2+ +2e-

reduction: NO3- + 4H+ + 3e- --> NO + 2H2O

The lowest possible number of moles of electrons is 6.

- Sat Feb 17, 2018 9:07 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: HW 14.117
- Replies:
**2** - Views:
**242**

### HW 14.117

HW 14.117: The body functions as a kind of fuel cell that uses oxygen from the air to oxidize glucose: C6H12O6 (aq) + 6O2 (g) -->6CO2 (g) + 6H2O (l) During normal activity, a person uses the equivalent of about 10 MJ of energy a day. Assume that this value represents delta G, and estimate the averag...

- Tue Feb 13, 2018 10:54 am
- Forum: Phase Changes & Related Calculations
- Topic: Practice Problem
- Replies:
**3** - Views:
**235**

### Re: Practice Problem

From the reaction, delta H = -106.9 kJ is per mole of PbO. You need to first find the number of moles of 49.7g PbO and then multiply that number by deltaH to get the total heat supplied to the water. Then you divide that amount of heat by the specific heat capacity of water (4.18J/g*°C) and deltaT (...

- Tue Feb 13, 2018 10:47 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Problem 9.69
- Replies:
**1** - Views:
**138**

### Re: Problem 9.69

You multiply delta G of the NADH reaction and the formation of H2O both by 3 and divide the sum by the delta G of the reaction for ATP. Note that e- form in the NADH reaction and you have to balance that using the H2O reaction.

- Tue Feb 13, 2018 10:42 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.81
- Replies:
**1** - Views:
**146**

### Re: 11.81

You could use ICE box to find the concentration of each chemical, but that requires a lot of work. Comparing the equilibrium constants is a much easier way.

- Tue Feb 06, 2018 10:38 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: constant pressure and constant volume
- Replies:
**3** - Views:
**166**

### Re: constant pressure and constant volume

For ideal gases's specific heat capacity:

At constant volume:

monatomic gas: 3/2R

diatomic gas: 5/2R

At constant pressure (=Cv+R):

monatomic gas: 5/2R

diatomic gas: 7/2R

At constant volume:

monatomic gas: 3/2R

diatomic gas: 5/2R

At constant pressure (=Cv+R):

monatomic gas: 5/2R

diatomic gas: 7/2R

- Tue Feb 06, 2018 10:27 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Anode vs. Cathode
- Replies:
**2** - Views:
**358**

### Re: Anode vs. Cathode

In general, the anode is the positive side and the cathode is the negative side. This refers to the movement of current. In electrochemistry, electrons flow the opposite way. The anode is the negative side and the cathode is the positive side.

- Tue Feb 06, 2018 10:22 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: G for non-equilibrium
- Replies:
**2** - Views:
**181**

### Re: G for non-equilibrium

Q is the reaction quotient. You use the same formula to calculate K (equilibrium constant) to calculate Q in a non-equilibrium state.

- Wed Jan 31, 2018 10:28 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.9
- Replies:
**3** - Views:
**196**

### Re: 9.9

For ideal gas, ΔU=3/2nRΔT

For isothermal reactions, ΔT=0, so ΔU also equals to 0.

For isothermal reactions, ΔT=0, so ΔU also equals to 0.

- Wed Jan 31, 2018 10:25 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: sign of entropy
- Replies:
**11** - Views:
**655**

### Re: sign of entropy

The sign of entropy indicates the direction of heat flow.

- Wed Jan 31, 2018 10:16 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Change in S with changes in P, V, and T
- Replies:
**3** - Views:
**247**

### Re: Change in S with changes in P, V, and T

It is not possible for all three to change simultaneously. When T changes, you use the formula with the heat capacity. The problem would tell you if the reaction is under constant P or V, then you would know which heat capacity to use.

- Wed Jan 24, 2018 10:54 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: C(deltaT) vs mC(deltaT)
- Replies:
**4** - Views:
**1042**

### Re: C(deltaT) vs mC(deltaT)

You use heat capacity when you don't have a total mass. That is if you know the total mass of the system, you can use the specific heat capacity. In this problem, you are given the heat capacity of the bomb calorimeter and change in temperature. So, you can get q by multiplying them together.

- Wed Jan 24, 2018 10:44 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.63
- Replies:
**4** - Views:
**266**

### Re: 8.63

The standard enthalpy of formation for H2O(l) should be -285.83 kJ/mol.

standard reaction enthalpy = [(-471.5 kJ/mol) * 1 mol + 2 mol * (-285.83 kJ/mol)] - [(-39.7 kJ/mol) * 1 mol + 2 mol * (-482.37 kJ/mol)]

=-38.72 kJ

standard reaction enthalpy = [(-471.5 kJ/mol) * 1 mol + 2 mol * (-285.83 kJ/mol)] - [(-39.7 kJ/mol) * 1 mol + 2 mol * (-482.37 kJ/mol)]

=-38.72 kJ

- Wed Jan 24, 2018 10:37 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Boltzmann's Equation in relation to temperature
- Replies:
**4** - Views:
**247**

### Re: Boltzmann's Equation in relation to temperature

The only factor in Boltzmann's entropy equation is microstate. In order for the arrangements of particles to give rise to the same total energy, the temperature has to be constant.

- Thu Jan 18, 2018 10:56 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Specific Heat Capacity
- Replies:
**5** - Views:
**1753**

### Re: Specific Heat Capacity

It is just a convention to assume the specific heat capacity of the solution is the same as the solvent.

- Thu Jan 18, 2018 10:40 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Delta H and q Interchangeability
- Replies:
**5** - Views:
**368**

### Re: Delta H and q Interchangeability

delta H only equals to q at constant pressure.

- Thu Jan 18, 2018 10:27 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Gas in a piston with regards to pressure
- Replies:
**3** - Views:
**190**

### Re: Gas in a piston with regards to pressure

Gas in a piston does 171kJ of work means that the volume increases. Using PV=nRT, the amount of gas and T stays the same (and R is a constant). Since V increases, the pressure should decrease to satisfy the equation.

- Wed Jan 10, 2018 10:18 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Equations
- Replies:
**6** - Views:
**359**

### Re: Equations

The most general equation is ΔH = H(final) - H(initial), and just plug in value for any final and initial state.

- Wed Jan 10, 2018 10:14 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Why does heat at constant pressure equal enthalpy?
- Replies:
**1** - Views:
**91**

### Re: Why does heat at constant pressure equal enthalpy?

By definition, H = U + PV, where U is internal energy, P is pressure of the system, V is volume. Also, ΔU = w + q, where q is heat and w is work. As a result, ΔH= ΔU + PΔV = w + q + PΔV, and w = -P(external) ΔV when you combine them together, ΔH = q - PexΔV + PΔV if pressure is constant, Pex = P, ΔH...

- Wed Jan 10, 2018 10:05 pm
- Topic: Question about today's lecture problem?
- Replies:
**3** - Views:
**177**

### Re: Question about today's lecture problem?

1. N2 (g) + O2 (g) -> 2NO (g) 2. 2NO (g) + O2 (g) -> 2NO2 (g) [it should be NO2 instead of NO] 3. N2 (g) + 2O2(g) -> 2NO2 (g) So, when you add up the first and the second reactions, you get N2(g) + O2(g) + 2NO (g) + O2(g) -> 2NO(g) + 2NO2(g). Then you combine O2 and cancel 2NO on both sides to get r...

- Tue Dec 05, 2017 5:21 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.57
- Replies:
**3** - Views:
**316**

### Re: 12.57

For part a, since you are given the pH value of HClO2, you can first calculate the molarity of H3O+. Then, you get the molarity of its conjugate base as it is the same as the hydronium ions. The question you probably have when calculating Ka is the equilibrium concentration of HClO2, which you get b...

- Tue Dec 05, 2017 5:14 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: HCOOH vs CH3COOH
- Replies:
**2** - Views:
**308**

### Re: HCOOH vs CH3COOH

Another explanation is that the CH3 group pushes electrons towards the O-H bond, making the bond stronger and thus the acid weaker. Therefore, CH3COOH is slightly weaker than HCOOH.

- Mon Nov 27, 2017 2:40 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Solids when using ICE [ENDORSED]
- Replies:
**8** - Views:
**1048**

### Re: Solids when using ICE [ENDORSED]

When using the ICE method, you ignore the concentrations for solids and pure liquids. But you use their stoichiometric coefficients to set up the equation.

- Mon Nov 27, 2017 2:14 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Chem Equilibrium Module 4 #15 [ENDORSED]
- Replies:
**6** - Views:
**754**

### Re: Chem Equilibrium Module 4 #15 [ENDORSED]

The liquids have no effect on the equilibrium constant K, not the equilibrium composition. Remember the equilibrium composition is the concentration of each chemical at equilibrium. In this reaction, water is the reactant. Therefore, when water is added, the reaction will shift to the right to minim...

- Tue Nov 21, 2017 4:28 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.13
- Replies:
**2** - Views:
**238**

### Re: 11.13

K is the equilibrium constant, while Q is the reaction quotient. K is determined only at chemical equilibrium, while Q can be evaluated at any stage. Only at equilibrium, Q=K. You can use the value of Q to see in which way will the reaction move towards.

- Tue Nov 21, 2017 4:20 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating Equilibrium Constant
- Replies:
**2** - Views:
**110**

### Re: Calculating Equilibrium Constant

You do not include the solids because you assume them to be pure. So, you do not really have a concentration for the solids (actually the number for it is just 1). Therefore, it is said to be not included in the equilibrium constant calculation.

- Sat Nov 18, 2017 4:33 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Polydentate ligands and shape
- Replies:
**1** - Views:
**171**

### Re: Polydentate ligands and shape

Yes, EDTA is a hexadentate ligand and it forms six coordinate bonds with Fe(III). Therefore, iron would have the coordination number of 6 because of the six bonds formed. However, I believe that EDTA and iron would form a chelate complex.

- Wed Nov 15, 2017 11:46 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Dipole Moment
- Replies:
**3** - Views:
**180**

### Re: Dipole Moment

The more electronegative atom will have a partial negative charge, and the less electronegative atom will have an equal but partial positive charge. When we draw an arrow to represent this, we draw it from the positive charge pointing to the negative charge (δ+ to δ-).

- Wed Nov 15, 2017 11:37 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Shape vs. Electron Arrangement
- Replies:
**2** - Views:
**171**

### Re: Shape vs. Electron Arrangement

The molecular shape depends on only the location of atoms, while the electron arrangement is determined by both atoms and lone pairs. The molecular shape is determined by the electron arrangement. When you want to find out the molecular shape, first find the electron arrangement and then take out th...

- Wed Nov 15, 2017 11:29 am
- Forum: Hybridization
- Topic: Diapole
- Replies:
**4** - Views:
**376**

### Re: Diapole

Dipole moment occurs when electrons are not shared equally. It is caused by the difference in electronegativity of atoms. A more electronegative atom pulls electrons more, thus having a partial negative charge, causing the other atom to have an equal but partial positive charge.

- Wed Nov 08, 2017 3:52 pm
- Forum: Octet Exceptions
- Topic: Expanded Octed
- Replies:
**2** - Views:
**292**

### Re: Expanded Octed

The central atom can have an expanded octet if it is in the third energy level or higher (starting from P, S, Cl). First draw the Lewis structure following the octet rule. If you can achieve a lower energy structure (formal charge closer to zero) with an expanded octet, then add bonds or electrons t...

- Wed Nov 08, 2017 3:37 pm
- Forum: Resonance Structures
- Topic: Formal Charge vs. Electronegativity
- Replies:
**1** - Views:
**315**

### Re: Formal Charge vs. Electronegativity

I would say the most stable one is the dominant structure in all three resonance structures, which is when the formal charge is as close as possible to zero. The other two structures exist because it obeys the octet rule and the electronegativity difference makes it possible.

- Wed Nov 01, 2017 12:07 pm
- Forum: Resonance Structures
- Topic: Delocalized
- Replies:
**7** - Views:
**547**

### Re: Delocalized

Because the actually resonance structure is a blend of all possible Lewis structures. So, the electrons in double bonds are not bound to one specific bond between two atoms but rather located among the bonds between the atoms throughout the molecule.

- Wed Nov 01, 2017 11:58 am
- Forum: Lewis Structures
- Topic: 3.33 Part C
- Replies:
**3** - Views:
**258**

### Re: 3.33 Part C

Fluorine cannot form a double bond with nitrogen, otherwise F would have an unpaired electron (forming a radical).

- Wed Nov 01, 2017 11:51 am
- Forum: Lewis Structures
- Topic: Elements that have extended octets
- Replies:
**3** - Views:
**254**

### Re: Elements that have extended octets

Any elements in period 3 and later can have an expanded octet.

- Tue Oct 24, 2017 12:15 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: chapter 2 Q29
- Replies:
**1** - Views:
**161**

### Re: chapter 2 Q29

For (b), it says n=4, l=2, and ml=-2. It shows a specific orbital in 4d-orbital. According to Pauli exclusion principle, no more than 2 electron can occupy any given orbital. So, only two electrons can have these three quantum number. The same thing for (d). Only two electrons can occupy a 3d-orbital.

- Tue Oct 24, 2017 10:24 am
- Forum: Quantum Numbers and The H-Atom
- Topic: 2.39
- Replies:
**4** - Views:
**392**

### Re: 2.39

An excited state can happen when you go from a an s-orbital to a p-orbital. But its definition is that an atom with electrons in energy states higher than predicted by the building-up principle is said to be in an excited state. In this case, carbon should have two electrons in different p-orbital w...

- Tue Oct 24, 2017 9:54 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Difference between orbital, shell, and subshell?
- Replies:
**2** - Views:
**1181**

### Re: Difference between orbital, shell, and subshell?

Orbital is a region in which there is a high probability (95%) of finding an electron in an atom. It is a general concept. An s-orbital is a spherical region; a p-orbial has two lobes; a d-orbital has four lobes (except one). Shell is all the orbitals of a given principal quantum number (n). For exa...

- Wed Oct 18, 2017 9:36 am
- Forum: Photoelectric Effect
- Topic: Number of Photons
- Replies:
**4** - Views:
**288**

### Re: Number of Photons

Yes, the number of photons increases as the intensity increases. This is the explanation for light with particle-like property. If the energy of the photons reaches the threshold energy, then increasing the intensity of light will increase the number of electrons ejected since it is an one-photon-on...

- Wed Oct 18, 2017 9:23 am
- Forum: Photoelectric Effect
- Topic: Post Module #29 (KJ*mol^-1 problem)
- Replies:
**4** - Views:
**293**

### Re: Post Module #29 (KJ*mol^-1 problem)

You're right. It does ask for the work function. However, the work function it gives you is in KJ.mol-1. So, you have you first multiply it by 10^3 to get J.mol-1. Then, you divide it by Avogadro constant to get J/atom.

- Tue Oct 10, 2017 10:34 pm
- Forum: Properties of Light
- Topic: Rydberg's Formula
- Replies:
**3** - Views:
**462**

### Re: Rydberg's Formula

In my opinion, we use the formula with the negative sign when we calculate the energy absorbed or emitted. That is the energy difference between two energy levels. We use the formula without the negative sign when we deal with the frequency associated with the light. Because energy emitted is negati...

- Tue Oct 10, 2017 10:15 pm
- Forum: Properties of Light
- Topic: Chapter 1 #3
- Replies:
**9** - Views:
**553**

### Re: Chapter 1 #3

As stated by the textbook, the electric field pushes the electron first in one direction and then pulls it in the opposite direction, over and over again. Electromagnetic radiation of frequency of 1 Hz pushes a charge in one direction, then the opposite direction, and returns to the original directi...

- Fri Oct 06, 2017 8:37 pm
- Forum: Limiting Reactant Calculations
- Topic: Extensive vs. Intensive properties
- Replies:
**2** - Views:
**379**

### Re: Extensive vs. Intensive properties

An extensive property depends on the size of the matter, such as volume. An intensive property doesn't depend on the amount, e.g. density. It doesn't change even though the size of the matter changes.

- Wed Oct 04, 2017 6:09 pm
- Forum: Empirical & Molecular Formulas
- Topic: Empirical and Molecular Formulas
- Replies:
**3** - Views:
**374**

### Re: Empirical and Molecular Formulas

The empirical formula is just the ratio of the atoms. The molecular formula shows the actual number of atoms. That is the molecular formula is the multiple of the empirical formula. In order to get the empirical formula from the molecular formula, you need to know the molar mass of the molecular for...

- Wed Oct 04, 2017 3:09 pm
- Forum: Significant Figures
- Topic: Theoretical and Percentage Yield
- Replies:
**4** - Views:
**451**

### Re: Theoretical and Percentage Yield

Theoretical yield is the maximum amount of a product you can get from a reaction. It is calculated from the mole ratio of a balanced equation. Percentage yield, on the other hand, is obtained from the following formula: (actual yield/theoretical yield)*100%. Some factors that may causes the differen...