## Search found 56 matches

- Sat Mar 17, 2018 2:19 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Entropy of vaporization? 9.37
- Replies:
**1** - Views:
**294**

### Re: Entropy of vaporization? 9.37

I think this is because the problem asks for the standard reaction entropy of the formation of H 2 O (l) from the elements in their most stable state at 298 K . Since you are calculating ΔS° using S° values, I don't think ΔS vap is included in the calculations. See the difference between the the ΔS°...

- Sat Mar 17, 2018 1:36 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Determining Slow Step
- Replies:
**2** - Views:
**2622**

### Re: Determining Slow Step

We did this in the reaction mechanism past exam problem in class on Wednesday. You would pretend every step is the slow step and find its rate law to see if it matches the overall rate law. It's a little bit of extra work to check every step but it's the same procedure as if you were told which step...

- Sat Mar 17, 2018 1:27 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation
- Replies:
**2** - Views:
**259**

### Re: Nernst Equation

E = E^{\circ} - \frac{RT}{nF}lnQ This is the general Nernst equation that can be used at any temperature (K). E = E^{\circ} - \frac{0.0592}{n}logQ This form of the Nernst equation is for calculations at 25°C (298 K). It was found by converting ln to log: ln = 2.503*log 10 Then using T=298 K: \frac{...

- Sat Mar 17, 2018 1:15 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Closed systems
- Replies:
**2** - Views:
**351**

### Re: Closed systems

Only closed systems will be affected by ΔU = q + w. Refer back to the definitions of open/closed/isolated systems: 1) open: both matter and energy can exchange with surroundings - can't measure ΔU because changes in matter are not accounted for in ΔU = q + w 2) closed: energy (q/w) can exchange with...

- Sun Mar 11, 2018 4:33 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Electrochem Review session problem 14.47
- Replies:
**2** - Views:
**202**

### Re: Electrochem Review session problem 14.47

Following the work in the solutions manual, Q = 3.3 x 10

^{6}, so technically your answer is the most accurate. I think the magnitude of Q is the most important point in the answer so that's why the solution manual says Q = 10^{6}.- Sun Mar 11, 2018 4:26 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.99
- Replies:
**1** - Views:
**147**

### Re: 9.99

The dehydrogenation of cyclohexane to benzene is not spontaneous: C 6 H 12 --> C 6 H 6 + 3H 2 ΔG° r = +97.6 kJ/mol However, if it is paired spontaneous process where a molecule can accept the hydrogen product, the entire process will turn spontaneous. Ethene accepting hydrogen is spontaneous: C 2 H ...

- Sun Mar 11, 2018 4:12 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.117
- Replies:
**1** - Views:
**152**

### Re: 8.117

In order to solve for ΔU, you need ΔH, P and ΔV. The given information says that ΔH = -318 kJ and that there is production of 1 mol of H 2 . One way to visualize this is to divide the entire equation by 3 to show that there is 1 mol of H 2 being produced: (1/3)CH 4 (g)+(1/3)H 2 O(g) --> (1/3)CO 2 (g...

- Sun Mar 04, 2018 8:56 pm
- Forum: Zero Order Reactions
- Topic: Initial concentrations
- Replies:
**2** - Views:
**231**

### Re: Initial concentrations

The integrated rate laws of zero, first and second order reactions depend on initial concentration because all of them contain [A] 0 . However, when deriving the half-life equations, only zero and second order reactions depend on initial concentration. 0 order reaction half-life: t 1/2 = [A] 0 /2k 1...

- Sun Mar 04, 2018 8:30 pm
- Forum: First Order Reactions
- Topic: 15.23 c
- Replies:
**2** - Views:
**224**

### Re: 15.23 c

First, find [A] t . The reaction starts with [A] 0 = 0.153 mol A and produces [B] t = 0.034 mol B during some time interval. Using the stoichiometric ratios in the given reaction, [A] 0 must have reacted at a rate of (2 mol A/1 mol B) to end up with [A] t = 0.153 mol/L A - (2 mol A/1 mol B)(0.034 mo...

- Sun Mar 04, 2018 8:20 pm
- Forum: General Rate Laws
- Topic: Half Life
- Replies:
**6** - Views:
**362**

### Re: Half Life

The half-life of a reaction is the time required for a reactant to reach half its original concentration. This information is helpful to predict the concentration of a substance over time. Another application that Dr. Lavelle mentioned in class is that knowing the half-life can help to identify an u...

- Fri Feb 23, 2018 2:26 pm
- Forum: Balancing Redox Reactions
- Topic: 14.17
- Replies:
**3** - Views:
**343**

### Re: 14.17

What happens in this reaction to the potassium and the chloride though? And are we supposed to assume that permanganate will always dissociate into Mn 2+ and Fe 2+ into Fe 3+ ? In this reaction and in most reactions, potassium ion and chloride are spectator ions because their oxidation states do no...

- Thu Feb 22, 2018 6:57 pm
- Forum: Balancing Redox Reactions
- Topic: 14.17
- Replies:
**3** - Views:
**343**

### Re: 14.17

Dr. Lavelle has this pdf ( https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Balancing_Redox_Reactions_Acidic_Conditions.pdf ) which explains how to get the half-reactions for this redox reaction. Follow each step to get the half-reaction for the iron and the half-reaction for the pe...

- Thu Feb 22, 2018 6:45 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell diagram
- Replies:
**4** - Views:
**313**

### Re: Cell diagram

There is no convention for which compound to put first, but it would make logical sense to place Cl

_{2}(g) before Cl-(aq) to show that Cl_{2}is being reduced to Cl-.- Wed Feb 14, 2018 2:31 pm
- Forum: Phase Changes & Related Calculations
- Topic: Correct Answer For Test 1 Question 7
- Replies:
**4** - Views:
**454**

### Re: Correct Answer For Test 1 Question 7

Yes, that is correct. The setup is q

_{ice}=-q_{tea}, which is nΔH_{fus}+ mCΔT = -mCΔT. You plugged in all the right numbers and got the correct final temperature.- Wed Feb 14, 2018 2:20 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: given two temperatures and 2 volumes solve for delta S
- Replies:
**2** - Views:
**371**

### Re: given two temperatures and 2 volumes solve for delta S

Since entropy is a state function, you can calculate ΔS in steps. In step 1, assume isothermal conditions and consider the change in volume to use ΔS 1 =nRln(V1/V2). Then in step 2, assume isochoric (no change in volume) conditions and consider the change in temperature to use ΔS 1 =nCln(T1/T2). In ...

- Wed Feb 14, 2018 2:15 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Delta U as 0
- Replies:
**3** - Views:
**313**

### Re: Delta U as 0

Those are all correct. What these conditions all have in common is that there is no change in temperature. This fact can be given if the problem says the process is explicitly isothermal, or no q is added/removed (adiabatic), or you could possibly deduct no change in temperature from PV=nRT.

- Tue Feb 13, 2018 2:03 pm
- Forum: Calculating Work of Expansion
- Topic: Practice Midterm 4
- Replies:
**3** - Views:
**318**

### Re: Practice Midterm 4

You have all the correct concepts and formulas. Check your units to see if you have made any errors. In the first step, you will use w = -PΔV but make sure you use R = 101.3 J/L*atm to get w = +3.04 kJ. As you stated, the work in the second step is zero. For the third step, use w = -nRTln(V2/V1) = -...

- Thu Feb 08, 2018 8:35 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.47
- Replies:
**2** - Views:
**201**

### Re: 9.47

You would use the initial volume because the problem states that the "ideal gas at 323 K occupies 1.67 L at 4.95 atm." This phrase describes the conditions of the ideal gas, and you would use these values in the ideal gas law to calculate an unknown value (which in this case is moles).

- Thu Feb 08, 2018 8:30 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.83
- Replies:
**2** - Views:
**183**

### Re: 11.83

That is also a viable method to solve this problem because you're solving for K at 25°C (standard state). If you look up the standard Gibbs free energy of formation for the products and reactants, you can use the difference of sums method to calculate ΔG°. Set ΔG°=-RTlnK to solve for K at 25°C. Then...

- Thu Feb 08, 2018 6:22 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.35
- Replies:
**1** - Views:
**153**

### Re: 9.35

0.5 mol is used for B and C because they are both described as diatomic molecules. This means that 1 mol of atoms is actually 0.5 mol of gas since the atoms pair up. 1 vibrational degree means that we are dealing with a "complex" molecule so you would use the corresponding Cv value of 3R. ...

- Wed Jan 31, 2018 10:56 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: HW 9.35 Vibrational degree of freedom
- Replies:
**1** - Views:
**177**

### Re: HW 9.35 Vibrational degree of freedom

I think that the 1 vibrational degree of freedom comes from the phrase "vibrationally active" given in the question. This means that the gas has additional vibrational states and is therefore more "complex" so you would use the 3R value of Cv for complex molecules. The diatomic m...

- Wed Jan 31, 2018 10:42 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.1 [ENDORSED]
- Replies:
**3** - Views:
**212**

### Re: 9.1 [ENDORSED]

So the signs would be in reference to the system? This particular problem asks for "the rate that your body generates entropy in your surroundings." Therefore, the answers would have signs in reference to the surroundings. The sign of ΔS depends on direction of heat flow. This example is ...

- Wed Jan 31, 2018 10:27 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: homework 9.19
- Replies:
**4** - Views:
**267**

### Re: homework 9.19

To find ΔS for change in temperature (for example: in the first step of this problem, T1=358 K and T2=373 K), you would use the equation for ΔS=n*C p *ln(T2/T1) which is derived from ΔS=q/T and nCΔT. Since this problem is asking for standard entropy, you can use the formula without n (ΔS=C p *ln(T2/...

- Fri Jan 26, 2018 2:12 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Calculating internal energy in isobaric reaction
- Replies:
**3** - Views:
**596**

### Re: Calculating internal energy in isobaric reaction

If we use the formula nCpDeltaT, where 5/2R can be replaced for Cp to calculate the change in internal energy for an ideal gas at constant pressure , how would we calculate the change in internal energy for just a normal isobaric reaction that did not involve a gas? For an isobaric process, you wou...

- Fri Jan 26, 2018 2:07 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Calculating internal energy in isobaric reaction
- Replies:
**3** - Views:
**596**

### Re: Calculating internal energy in isobaric reaction

Dylan Mai 1D wrote:Can someone also explain what an isobaric reaction is?

An isobaric process occurs when the pressure remains constant. To calculate work in an isobaric process, you would use the equation w= -P

_{ex}ΔV.

- Wed Jan 24, 2018 1:09 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Lecture 1/24 (Wednesday) [ENDORSED]
- Replies:
**5** - Views:
**250**

### Re: Lecture 1/24 (Wednesday) [ENDORSED]

The calculation of a reversible, isothermal expansion of a gas is explained on pages 265-266 of Chapter 8 in the textbook. The green shaded box shows the derivation of the formula using integrals, which is what Dr. Lavelle did in class today.

- Sat Jan 20, 2018 4:42 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Example 8.7
- Replies:
**1** - Views:
**105**

### Re: Example 8.7

The units for ΔH are Joules. If you are trying to find the change in enthalpy (ΔH) for 2 mol of benzene, you need to multiply by q (J) in the numerator and n (mol) in the denominator to end up with the ΔH in Joules. \Delta H = \frac{(2 mol)\cdot (-8.60\cdot 551J)}{(0.113/78.12 mo...

- Sat Jan 20, 2018 4:30 pm
- Forum: Phase Changes & Related Calculations
- Topic: HW #1 PART E
- Replies:
**3** - Views:
**283**

### Re: HW #1 PART E

A closed system is one that can exchange energy but not matter with the surroundings. Mercury in a thermometer exchanges energy with its surroundings to rise and fall in the tube and indicate the temperature, but it does not leak and exchange matter. Hope this helps!

- Sat Jan 20, 2018 4:21 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Specific Heat Capacity
- Replies:
**3** - Views:
**239**

### Re: Specific Heat Capacity

Yes, K stands for Kelvin. The "steps" or the magnitude of degrees on the Kelvin and Celsius scale are equivalent so they can be used interchangeably in definitions like units.

- Fri Jan 12, 2018 3:24 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.25
- Replies:
**1** - Views:
**130**

### Re: 8.25

Since the question states that this is occurring in a constant-volume calorimeter, you don't need to worry about volumes. In addition, the heat capacity of a calorimeter is an extensive property so moles don't matter. Using q=CΔT rearranged to C=q/ΔT, you can calculate this heat capacity and then us...

- Fri Jan 12, 2018 3:13 pm
- Forum: Phase Changes & Related Calculations
- Topic: Enthalpy of vaporization
- Replies:
**2** - Views:
**219**

### Re: Enthalpy of vaporization

Enthalpy of vaporization (ΔH vap ) is the difference in molar enthalpy (H m ) between the vapor and liquid states. All the "H"s mean enthalpy. The "Δ" in ΔH vap shows the change in enthalpy of the vapor and liquid states. The "m" in H m indicates that we are using molar...

- Fri Jan 12, 2018 1:43 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: heat capacity of a gas at constant pressure
- Replies:
**1** - Views:
**108**

### Re: heat capacity of a gas at constant pressure

Usually the C

_{p}or C_{v}will be given in the question unless there is sufficient information (q, n/m, ΔT) given to calculate the missing C. For a complete table of C_{p}values, you can refer to the textbook Appendix 2A.- Sat Dec 09, 2017 10:49 pm
- Forum: Conjugate Acids & Bases
- Topic: % Ionization [ENDORSED]
- Replies:
**3** - Views:
**605**

### Re: % Ionization [ENDORSED]

For a given equation: acid + H 2 O ⇌ H 3 O + + conjugate base % ionization/dissociation = ([equilibrium concentration of H 3 O + ] / [initial concentration of acid]) x 100 For a given equation: base + H 2 O ⇌ OH - + conjugate acid % ionization/dissociation = ([equilibrium concentration of OH - ] / [...

- Sat Dec 09, 2017 10:39 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Spin Quantum Number: Test 3
- Replies:
**3** - Views:
**333**

### Re: Spin Quantum Number: Test 3

The question asks "Write all the possible values that a spin quantum number can take for this electron." It can be either +1/2 or -1/2 because the "positive" and "negative" directions are relative. The + and - do not necessarily mean "up" or "down;" ...

- Sun Dec 03, 2017 3:13 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Sig Figs for Calculating pH and pOH
- Replies:
**4** - Views:
**398**

### Sig Figs for Calculating pH and pOH

For calculating logarithms/pH/pOH, the number of sig figs comes from the mantissa aka after the decimal place. However, for some problems (such as 12.27 and 12.29), the solutions manual does not acknowledge this. If the calculation was pH=-log(0.0 25 ) (2 sig figs), wouldn't the answer be 1. 60 with...

- Sun Dec 03, 2017 12:46 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.35
- Replies:
**1** - Views:
**153**

### Re: 12.35

Yes, you would solve for K

_{a2}using the same antilog method. K_{a2}is the disassociation constant for the second deprotonation for polyprotic acids, while K_{a1}is the disassociation constant for the first deprotonation. In this problem, since you are given pK_{a2}, you would solve for K_{a2}.- Thu Nov 30, 2017 2:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.11 part C
- Replies:
**1** - Views:
**152**

### Re: 11.11 part C

The ratio of [O 2 ]/[O 3 ] is different from [O 2 ] 3 /[O 3 ] 2 because the concentrations are raised to different powers. If [O 2 ]=2 and [O 3 ]=3, [O 2 ]/[O 3 ] would be 2/3 and [O 2 ] 3 /[O 3 ] 2 would be (2) 3 /(3) 2 =8/9. You can see that 2/3 does not equal 8/9, so [O 2 ]/[O 3 ] does not equal ...

- Thu Nov 30, 2017 2:06 pm
- Forum: Lewis Structures
- Topic: Lewis Structure of IO2F2 (-2) Problem 4.23 [ENDORSED]
- Replies:
**3** - Views:
**4971**

### Re: Lewis Structure of IO2F2 (-2) Problem 4.23 [ENDORSED]

I did the same thing to create a more stable structure. The solution manual notes that other Lewis structures with double bonds are possible, but the use of single vs. double bonds does not change the shape of the molecule. If you draw it with single or double bonds, the molecule is still see-saw, s...

- Tue Nov 21, 2017 1:00 pm
- Forum: Hybridization
- Topic: 4.31
- Replies:
**2** - Views:
**167**

### Re: 4.31

Yes, that's exactly how. The hybridization is based on the areas of electron density around the atom. In other words, it's based on how many "things" (atoms or lone pairs) are attached to the central atom (or the particular atom you are looking at - every atom in the molecule can have hybr...

- Tue Nov 21, 2017 12:52 pm
- Forum: Hybridization
- Topic: 4.73
- Replies:
**2** - Views:
**191**

### Re: 4.73

A molecule is considered a radical if it has an odd number of electrons. CH 2 has 6 total electrons and CH 2 2+ has 4 total electrons. The incomplete octet of the central atom does not directly affect whether a molecule is a radical. However, if carbon had an incomplete octet of with an odd number o...

- Thu Nov 16, 2017 5:44 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: Bond Angles
- Replies:
**8** - Views:
**721**

### Re: Bond Angles

You can just memorize/visualize the bond angles for each VSEPR shape and then if there are any lone pairs present (AX 2 E , AX 3 E 2 , etc.), the bond angle would be less than the "original" bond angles (listed below). You can just state that the bond angles are <___° (the bond angles with...

- Wed Nov 15, 2017 10:42 pm
- Forum: Hybridization
- Topic: Order of s,p,d in Names of Hybrid Orbitals
- Replies:
**3** - Views:
**329**

### Order of s,p,d in Names of Hybrid Orbitals

When naming hybridization, do the hybrid orbitals have to be written in the same order as the atomic orbitals (s,p,d,f) in electron configurations? For example, when we write the electron configuration of elements that contain electrons in the d-orbital, we write Ni: [Ar]3d 8 4s 2 , with the d-orbit...

- Tue Nov 07, 2017 2:39 pm
- Forum: Lewis Structures
- Topic: 3.41
- Replies:
**1** - Views:
**285**

### Re: 3.41

The formula for glycine is H2C(NH2)COOH. The NH2 in parentheses after the first carbon means that NH2 is attached to the carbon in H2C. So from left to right, draw NH2 and then attach the nitrogen to the carbon in H2C. Then, add COOH by attaching a carbon to the carbon in H2C. Add the oxygens and hy...

- Mon Nov 06, 2017 4:43 pm
- Forum: Dipole Moments
- Topic: Dipole Moment of CO
- Replies:
**3** - Views:
**1603**

### Re: Dipole Moment of CO

The difference in electronegativity between carbon and oxygen creates a small dipole moment. Only diatomic atoms (like O2) are truly covalent.

- Wed Nov 01, 2017 9:51 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Exceptions to Electron Configurations
- Replies:
**1** - Views:
**172**

### Re: Exceptions to Electron Configurations

I'm also curious as to why tungsten is not an exception like chromium and copper. From my knowledge, the exceptions with special configurations are chromium and molybdenum (s1d5) and copper, silver and gold (s1d10).

- Wed Nov 01, 2017 9:42 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Order of Orbitals
- Replies:
**10** - Views:
**1660**

### Re: Order of Orbitals

The order of orbitals does matter. When you write the electron configuration for an element, you must follow the Aufbau ("building up") Principle and write the orbitals in order of increasing energy. Therefore, if you have 3d and 4s, you need to write 3d before 4s because the 3d state is l...

- Wed Nov 01, 2017 9:37 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Copper and Chromium [ENDORSED]
- Replies:
**4** - Views:
**748**

### Re: Copper and Chromium [ENDORSED]

These configurations for copper and chromium are more stable because they result in a more symmetric distribution of the electrons. For chromium, the electron configuration [Ar]3d5 4s1 distributes the 6 valence electrons evenly. 5 electrons fill the 5 d-orbitals and 1 electron fills the single s-orb...

- Wed Oct 25, 2017 3:39 pm
- Forum: Trends in The Periodic Table
- Topic: List of all trends?
- Replies:
**5** - Views:
**637**

### Re: List of all trends?

I find it helpful to draw two diagonal arrows (one towards the top right and the other towards the bottom left corner) and label the trends. Increase across a period and up a group (towards top right corner): ionization energy, electron affinity, nonmetallic character Increase left on a period and d...

- Wed Oct 25, 2017 3:31 pm
- Forum: Trends in The Periodic Table
- Topic: Problem 2.67 (c) and (d)
- Replies:
**2** - Views:
**217**

### Re: Problem 2.67 (c) and (d)

In addition to increasing across a period, electron affinity values generally increase up a group. This is seen in the two problems posted. Chlorine is above bromine and has a higher electron affinity. Lithium is above sodium so it also has a higher electron affinity.

- Tue Oct 17, 2017 10:19 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Question about Atomic Spectroscopy
- Replies:
**4** - Views:
**372**

### Re: Question about Atomic Spectroscopy

wait so if every element has a unique spectroscopy then how do our problems all coincide with the same spectrums/ do we need to memorize the spectrum lines (Laymen series etc)??? All our homework/test problems concerning spectroscopy deal with hydrogen, so we have only been working with its atomic ...

- Tue Oct 17, 2017 10:01 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Question about energy levels [ENDORSED]
- Replies:
**8** - Views:
**619**

### Re: Question about energy levels [ENDORSED]

There is no specific cut off for the highest energy level, but most elements have energy levels from n=1 to about n=6-7. This is because n=1-7 is the range of quantum numbers seen in the periodic table of elements (7 rows = 7 energy levels). Most diagrams draw the energy levels up to n=6-7 and then ...

- Fri Oct 13, 2017 3:12 pm
- Forum: Properties of Light
- Topic: Exercise 1.9 Sig Fig
- Replies:
**3** - Views:
**297**

### Re: Exercise 1.9 Sig Fig

I think that technically the answer would be 6.0 x 10^-7 m for wavelength. However, it is easier to identify and compare wavelengths if the value is given in nanometers so it would be converted to 600 nm. If you wanted to be specific about sig figs, you could write a bar above the first zero to indi...

- Fri Oct 13, 2017 2:53 pm
- Forum: Properties of Light
- Topic: Heisenberg uncertainty Principle
- Replies:
**1** - Views:
**182**

### Re: Heisenberg uncertainty Principle

Heisenberg's uncertainty principle states that you can never simultaneously know the exact position and the exact speed/momentum of an object. This is expressed in the equation ΔpΔx ≥ (1/2)(h bar). Δx is the uncertainty in position and Δp is the uncertainty in momentum and the product of these must ...

- Wed Oct 04, 2017 8:23 pm
- Forum: Significant Figures
- Topic: E29 (d) Significant Figures
- Replies:
**3** - Views:
**405**

### Re: E29 (d) Significant Figures

In this question, the value 8.61 g is actually not relevant. Part (d) asks "What fraction of the total mass of the sample was due to oxygen?" All the values you need for this are derived from the periodic table, not the 8.61 at the beginning of the question. To find the fraction of the mas...

- Wed Oct 04, 2017 8:13 pm
- Forum: SI Units, Unit Conversions
- Topic: Limiting Reagents [ENDORSED]
- Replies:
**8** - Views:
**709**

### Re: Limiting Reagents [ENDORSED]

Generally, if the problem gives you an amount (in grams or moles) of each reactant to make a calculation, you are implicitly asked to find the limiting reactant. A good example of this is question M.7 (b), which asks "What mass of boron can be produced when 125 kg of boron oxide is heated with ...

- Wed Oct 04, 2017 8:05 pm
- Forum: SI Units, Unit Conversions
- Topic: Exercise L39 part b
- Replies:
**2** - Views:
**234**

### Re: Exercise L39 part b

SnO2 is an ionic compound so you would use the procedure to write the name for ionic compounds (metal + nonmetal), which is: 1. Determine the charge on the cation (Sn = +4) and anion (O = -2). 2. Name the cation. If it is a transition metal, put its charge in parentheses. For this example, you would...