Search found 55 matches

by miznaakbar
Sun Mar 18, 2018 3:20 pm
Forum: Student Social/Study Group
Topic: Post All Chemistry Jokes Here
Replies: 9651
Views: 3641840

Re: Post All Chemistry Jokes Here

Hey, do you have 11 protons?

Because you're sodium fine
by miznaakbar
Sun Mar 18, 2018 3:17 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat Capacity of Water
Replies: 4
Views: 778

Re: Heat Capacity of Water

So based off how water's heat capacity is dependent upon its hydrogen bonds, does that mean heat capacity for other substances is dependent upon their intermolecular forces? Or are there other factors that affect heat capacity?
by miznaakbar
Sun Mar 18, 2018 3:08 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Adiabatic System
Replies: 5
Views: 898

Re: Adiabatic System

Even if you assume it is adiabatic and q=0, you cannot figure out whether it is reversible or irreversible without more information, such as if pressure or volume is constant or not. It also depends on whether the system is closed, open, or isolated.
by miznaakbar
Sun Mar 18, 2018 3:04 pm
Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
Topic: Electrolytic cells vs Galvanic Cells
Replies: 3
Views: 1115

Re: Electrolytic cells vs Galvanic Cells

When given an electrolytic cell, make sure to set up your redox reactions so that you get a negative overall standard reduction potential. But when given an galvanic cell, make sure to get a positive overall standard reduction potential. The reason galvanic cells, which have spontaneous redox reacti...
by miznaakbar
Sun Mar 11, 2018 6:02 pm
Forum: Second Order Reactions
Topic: Half life of second order reactions
Replies: 8
Views: 1335

Re: Half life of second order reactions

I'm not sure if they've asked to derive a half life formula before, but it would probably be best to know how to considering we spent time deriving formulas in the notes. Deriving the half life formula is also fairly simple when given the integrated rate law because you just plug in (1/2)[A] initial...
by miznaakbar
Sun Mar 11, 2018 5:54 pm
Forum: Balancing Redox Reactions
Topic: redox
Replies: 6
Views: 917

Re: redox

The oxidizing agent is the compound that is reduced, or gains electrons. The reducing agent is the compound that is oxidized, or loses electrons.
by miznaakbar
Sun Mar 11, 2018 3:31 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Heat Required to Heat a Substance
Replies: 5
Views: 701

Re: Heat Required to Heat a Substance

The question seems to be referring to the specific/molar heat capacity of these two substances, so it is asking which substance requires more heat to raise 1 mol by 1 degree Celsius. You would need to look at a table of values of specific and molar heat capacities to determine whether water or butan...
by miznaakbar
Sun Mar 04, 2018 12:35 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: K'
Replies: 7
Views: 917

Re: K'

We use k' for pseudo order rate laws and it denotes the rate constant at the specific conditions you are studying. For psuedo order rate laws, you generally obtain k' from the slope of the linear plots you are looking it. You can't use the formulas on the sheet because you're finding k at altered co...
by miznaakbar
Sun Mar 04, 2018 12:29 pm
Forum: First Order Reactions
Topic: Pseudo-1st-Order
Replies: 6
Views: 821

Re: Pseudo-1st-Order

When working with pseudo-1st order rate laws, it's also important to note that the order(n) is usually determined by looking at linear plots of [A] vs t (n=0), ln([A]) vs t (n=1), and (1/[A]) vs t (n=2). Whichever one is linear tells you which order the reaction is.
by miznaakbar
Sun Mar 04, 2018 12:24 pm
Forum: General Rate Laws
Topic: 15.39a
Replies: 3
Views: 475

Re: 15.39a

In lecture when we derived the differential rate law for second order reactions, we assumed a=1. For our purposes, it seems like they want us to assume that the coefficients do not affect the reaction (like the solutions manual shows), though I'm not really sure why they would not want us to take co...
by miznaakbar
Sun Feb 25, 2018 4:42 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Using K
Replies: 3
Views: 504

Re: Using K

Which equation are you referring to?
by miznaakbar
Sun Feb 25, 2018 4:39 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Kinetic vs. Thermodynamics
Replies: 3
Views: 544

Re: Kinetic vs. Thermodynamics

To add to that, kinetics addresses the actual process/path of a reaction, which includes the rate, activation energy etc. Meanwhile, thermodynamics addresses the before and after states of a reaction, such as the change in free energy, spontaneity (whether or not a rxn will occur). So thermodynamics...
by miznaakbar
Sun Feb 25, 2018 4:33 pm
Forum: General Rate Laws
Topic: Gas Product
Replies: 3
Views: 514

Re: Gas Product

Another way to think of it is that N2 leaving the solution as a gas reduces the amount of reactant available for the reverse reaction. N2 is acting as a limiting reactant here, so the H2O does not have enough available N2 to react with in order to form more NO2 and NH4. Hope this helps
by miznaakbar
Fri Feb 23, 2018 5:09 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: The rate of a reaction
Replies: 5
Views: 768

Re: The rate of a reaction

Yes, rates of reactions is all dependent on kinetics, which includes the nature of the reactants, the temperature, the presence of a catalyst, and concentration. Thermodynamics is what we use to see whether or not reactions are favorable, not their speed.
by miznaakbar
Sat Feb 17, 2018 7:47 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Negative E
Replies: 2
Views: 691

Re: Negative E

In addition, when E˚(the standard reduction potential) is negative, the species tends to be a reducing agent because the more negative E˚is, the more willing it is to give up electrons, or be oxidized.
by miznaakbar
Sat Feb 17, 2018 7:10 pm
Forum: Balancing Redox Reactions
Topic: Oxidizing & Reducing Agents
Replies: 6
Views: 1184

Re: Oxidizing & Reducing Agents

Oxidizing agents are the species being reduced, or gaining electrons. Oxidizing agents cause another species to go through oxidation, or electron loss; the oxidizing agent thus gains the electrons the other species lost (which is why it is reduced) Reducing agents are the species being oxidized, or ...
by miznaakbar
Sat Feb 17, 2018 7:06 pm
Forum: Balancing Redox Reactions
Topic: Acidic vs. Basic Solutions
Replies: 3
Views: 517

Re: Acidic vs. Basic Solutions

According to the book, after you've balanced all other elements besides O and H, you do the following: For acidic: balance O using H2O, then balance H using H+ For basic: balance O using H2O, then balance H by adding H2O to the side that needs H and adding OH- to the other side (by adding H2O to one...
by miznaakbar
Sun Feb 11, 2018 11:28 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Test #7
Replies: 7
Views: 905

Re: Test #7

To specify my earlier response, the equation should look like this, where Temp final is unknown: (melt ice) + (raise water at 0 degrees C to temp final) = - (decreasing tea water at 30 degrees C to temp final) we want q(ice cube) = - q (tea water), so the negative sign is crucial: [(Enthalpy of Fusi...
by miznaakbar
Sun Feb 11, 2018 1:58 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Pressure Change
Replies: 1
Views: 234

Re: Pressure Change

Since PV=nRT, P and V are inversely related. The entropy equation with pressure is derived from the volume equation, which is why the initial and final values flip places so that the initial pressure term is above the final pressure term.
by miznaakbar
Sun Feb 11, 2018 1:45 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Test #7
Replies: 7
Views: 905

Re: Test #7

You need to multiply delta H (fusion) by the moles of ice you have so that you account for the energy required to change phases from ice to water. You would then use the specific heat of liquid water in calculations of q, not ice, since you have already accounted for its phase change in your calcula...
by miznaakbar
Sun Feb 11, 2018 1:42 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: "Ideal"
Replies: 6
Views: 773

Re: "Ideal"

Since an ideal system does not have to be under constant volume or pressure, does that mean that it does not matter whether it is open, closed, or isolated?
by miznaakbar
Sat Feb 03, 2018 7:57 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Test Number 1
Replies: 9
Views: 1083

Re: Test Number 1

The question stated: A balloon filled with gas is expanded to a larger volume and lower pressure while maintaining the same temperature. Answer yes or no if heat is being transferred in the process. My TA said that you do expansion work against the constant atmospheric pressure, which increases the ...
by miznaakbar
Sat Feb 03, 2018 7:53 pm
Forum: Phase Changes & Related Calculations
Topic: Degeneracy
Replies: 4
Views: 638

Re: Degeneracy

Degeneracy, denoted by W, is the number of ways of achieving a given energy state. We use degeneracy (W) in the Boltzmann equation, which is S = k x ln(W).
by miznaakbar
Sat Feb 03, 2018 7:47 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Delta G [ENDORSED]
Replies: 4
Views: 504

Re: Delta G [ENDORSED]

Delta G isn't always negative because you calculate it using enthalpy and entropy (which are not always negative values). For instance, if delta H was a positive value and delta S was a small positive values, then delta G would be a positive value. Delta G being negative indicates that a reaction is...
by miznaakbar
Sun Jan 28, 2018 1:51 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Gas Expansion [ENDORSED]
Replies: 3
Views: 522

Re: Gas Expansion [ENDORSED]

A gas would want to reach its maximum entropy because doing so maximizes the number of positions/states it can occupy. Thus, even without any change in heat, work, or internal energy, the gas will spontaneously expand (volume increase) because it is the most favorable possible state. Hope this helps.
by miznaakbar
Sun Jan 28, 2018 1:24 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Internal Energy and Spontaneity
Replies: 3
Views: 342

Re: Internal Energy and Spontaneity

Spontaneously, a gas wants to occupy the greatest number of positions/states, meaning the entropy increases. Thus, the gas expands its volume to reach its maximum entropy without any change in heat, work, or internal energy.
by miznaakbar
Sun Jan 28, 2018 1:16 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Homework 8.49
Replies: 4
Views: 927

Re: Homework 8.49

Since the question doesn't specify, we can assume it is at room temperature (which is 298K).
by miznaakbar
Fri Jan 19, 2018 12:09 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.67
Replies: 3
Views: 233

Re: 8.67

From what I understand, we should probably memorize the most stable states of common compounds and elements such as carbon. Otherwise you would just need to guess which state the product is, though it may often be easy to assume; for example in part a, both H2 and O2 are in the gas phase, so it woul...
by miznaakbar
Fri Jan 19, 2018 12:04 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Q 8.67
Replies: 1
Views: 159

Re: Q 8.67

For this problem, we use bond enthalpies to find the enthalpy of formation of the given compounds in their liquid state. For H2O, I wrote the balanced rxn as H2(g) + (1/2)O2(g) -> H2O(g) and then another phase change rxn to find the enthalpy for the liquid state: H2O(g)->H2O(l). For the first rxn fo...
by miznaakbar
Thu Jan 18, 2018 11:58 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: hw Q8.61
Replies: 1
Views: 2330

Re: hw Q8.61

From what I understand, rearranging is essentially trial and error, where you are trying to cancel out compounds that you do not want in your final equation through flipping equations and multiplying coefficients. First you would flip the 1st and 2nd rxns and write them in reverse so that the H2 and...
by miznaakbar
Fri Jan 12, 2018 12:54 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Elements in their most stable form
Replies: 2
Views: 11703

Re: Elements in their most stable form

From what I understand, you cannot assume that they are in their most stable state. You will need to know by looking at a chemical reaction whether or not, for instance, O2 as a gas is its most stable state. If any element is not in its stable form, you will need to write a phase change equation and...
by miznaakbar
Fri Jan 12, 2018 10:18 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Most Stable/Pure Form
Replies: 2
Views: 188

Re: Most Stable/Pure Form

From my understanding, purity is referring to there being only one element in the sample. So, for a sample to be in its stable state, it would need to be in its purest form. For instance, there could only be Oxygen in a gas sample (no other gases) for one to claim the sample is in its most stable st...
by miznaakbar
Fri Jan 12, 2018 10:11 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard Enthalpies of Formation of Diatomic Molecules
Replies: 6
Views: 4896

Re: Standard Enthalpies of Formation of Diatomic Molecules

No this rule applies to all elements, because in their standard state they are in their most stable state, so enthalpy is zero. For instance, Carbon's standard state is a solid as graphite, so it's enthalpy would be zero. However, Carbon in its solid state as a diamond would have a nonzero enthalpy ...
by miznaakbar
Sun Dec 10, 2017 4:29 pm
Forum: Properties & Structures of Inorganic & Organic Bases
Topic: NaOH [ENDORSED]
Replies: 1
Views: 418

Re: NaOH [ENDORSED]

NaOH is a strong base since it dissociates completely. It's often used in neutralization reactions with strong acids such as HCl to form water.
by miznaakbar
Sun Dec 10, 2017 4:27 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: example problem for Ka and Kb [ENDORSED]
Replies: 1
Views: 425

Re: example problem for Ka and Kb [ENDORSED]

You can first find pKa by doing 14 - pKb and then you can convert pKa into Ka, which is what you would then use to solve equilibria problems for the acid.
by miznaakbar
Sat Dec 02, 2017 5:13 pm
Forum: Amphoteric Compounds
Topic: Amphoteric and amphiprotic compounds
Replies: 4
Views: 634

Re: Amphoteric and amphiprotic compounds

Amphoteric substances can act as either an acid or a base whereas an amphiprotic substance can donate and accept hydrogen ions. Some but not all amphoteric substances are amphiprotic, for example page 429 of the textbook explains this well: "Distinguish between amphoteric and amphiprotic. Alumi...
by miznaakbar
Sat Dec 02, 2017 5:07 pm
Forum: Calculating the pH of Salt Solutions
Topic: Calculating pH
Replies: 3
Views: 517

Re: Calculating pH

From what i understand, the negative sign is just a way to ensure the pH is always a positive value. Generally, the concentration of ions is very small, so the value will be raised to a negative exponent. For instance, if you have a concentration of 1.0 x 10^-7, taking the negative log will give you...
by miznaakbar
Sat Nov 25, 2017 4:28 pm
Forum: Hybridization
Topic: Pi bonds and Sigma bonds
Replies: 1
Views: 464

Re: Pi bonds and Sigma bonds

For ethyne, each carbon is bonded to two atoms, therefore it will have two hybridized orbitals. This will lead to the formation of two sp hybridized orbitals and leaves 2 regular p orbitals. Now when we draw the orbitals, we will have two sp orbitals sticking out of the sides of each C atom. Both th...
by miznaakbar
Sat Nov 25, 2017 4:09 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sigma bonds versus Pi bonds
Replies: 1
Views: 870

Re: Sigma bonds versus Pi bonds

A sigma bond forms when two hybridized orbitals overlap (such as sp, sp2, or sp3 hybridized orbitals) whereas pi bonds form when 2 p orbitals overlap side by side. From what I understand, when looking at a lewis structure, a single bond will be a sigma bond, a double bond will have 1 pi and 1 sigma ...
by miznaakbar
Sun Nov 19, 2017 10:10 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Why lone pairs occupy equatorial plane
Replies: 2
Views: 3150

Re: Why lone pairs occupy equatorial plane

From what I understand, the explanation above is correct because, in simpler terms, the lone pairs want to be as far away from other regions of electron density as possible since it is more energetically favorable. If the lone pair is on the axial plane, it will be 90 degrees away from 3 regions of ...
by miznaakbar
Sun Nov 19, 2017 9:57 pm
Forum: Resonance Structures
Topic: Midterm Q8
Replies: 2
Views: 282

Re: Midterm Q8

HOCO is actually supposed to be a radical, and the single electron lies on the Carbon. For the structure of HOCO, you would have a single bond between H and O, a single bond between O and C, and then a double bond between the C and final O. You need to test out different combinations of double and s...
by miznaakbar
Fri Nov 10, 2017 11:12 am
Forum: Trends in The Periodic Table
Topic: Test 3 #3
Replies: 2
Views: 319

Re: Test 3 #3

Phosphorus, like Nitrogen, has a half filled p-subshell, which is stable so neither of these wants an electron as much as Oxygen or Sulfur. Thus, P will have a lower e affinity than S.
by miznaakbar
Fri Nov 10, 2017 11:03 am
Forum: Lewis Structures
Topic: Radicals in structures [ENDORSED]
Replies: 3
Views: 602

Re: Radicals in structures [ENDORSED]

From what I understand, the radical generally goes to the least electronegative atom (for example, in ch 3 question 123 the lewis structure of HOCO places the radical on the Carbon rather than one of the Oxygens).
by miznaakbar
Sun Nov 05, 2017 10:34 am
Forum: Lewis Structures
Topic: 3.57
Replies: 2
Views: 419

3.57

In the question, it asks to draw the lewis structures for the sulfite ion, hydrogen sulfite ion, perchlorate ion, and nitrite ion - for this class and the midterm, are we expected to know the structure of these ions just by looking at their name?
by miznaakbar
Sun Nov 05, 2017 10:22 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 3.9
Replies: 2
Views: 328

Re: 3.9

From what I understand, when an electron configuration includes electrons in both the d and s orbitals, such as [Ar] 3d^7 4s^2, the 4s level is higher in energy and thus electrons must be removed from it first. Therefore, when in question 3.9 it asks for which M^(2+) ions have the following electron...
by miznaakbar
Sun Oct 29, 2017 4:38 pm
Forum: Trends in The Periodic Table
Topic: Ionization Energy [ENDORSED]
Replies: 5
Views: 767

Re: Ionization Energy [ENDORSED]

Ionization energy decreases as you go down a group because as you go down a group, you add another shell of electrons. This makes the valence electrons further away from the nucleus' positive charge, which makes the electron easier to remove (thus lower ionization energy, or energy required to move ...
by miznaakbar
Sun Oct 29, 2017 4:33 pm
Forum: Trends in The Periodic Table
Topic: Diagonal Relationships of the Periodic Table [ENDORSED]
Replies: 4
Views: 1970

Re: Diagonal Relationships of the Periodic Table [ENDORSED]

Number 71 in the assigned ch 2 homework problems asks about diagonal relationships, and it simply means the similarity in chemical properties between an element and the element lying one period lower and one group to the right. For instance, similarities between the lithium ion (+1) and magnesium io...
by miznaakbar
Thu Oct 19, 2017 11:07 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Rydberg Equation
Replies: 7
Views: 898

Re: Rydberg Equation

My TA said it was all right to use the smaller n-value given for the n1 value, that way you will always get a positive number for your value of frequency (if you do not include the negative sign in front of R ). I usually just write a note on the side indicating which is n-initial and n-final.
by miznaakbar
Thu Oct 19, 2017 10:55 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Balmer and Lyman Wavelengths [ENDORSED]
Replies: 4
Views: 534

Re: Balmer and Lyman Wavelengths [ENDORSED]

^Looks like we're all on the same wavelength then
by miznaakbar
Thu Oct 19, 2017 10:51 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Balmer and Lyman Wavelengths [ENDORSED]
Replies: 4
Views: 534

Re: Balmer and Lyman Wavelengths [ENDORSED]

One of the homework questions said UV light was released from a transition, and it implied that it was a Lyman series (meaning it jumped down to n=1). Using n=1, I got the right answer so I assume that there are no exceptions considering the question expected you to know UV meant Lyman.
by miznaakbar
Fri Oct 13, 2017 11:45 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: lines in electromagnetic spectrum
Replies: 3
Views: 596

Re: lines in electromagnetic spectrum

I was pretty confused about that as well but I found this video that simplified the concept for me, here's the link:
https://www.youtube.com/watch?v=6rHerkru60E

If the link doesn't work try searching "2.2 The Line Spectrum of Hydrogen [SL IB Chemistry]" on youtube - hope this helps
by miznaakbar
Fri Oct 13, 2017 11:33 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Atomic spectra pre assessment
Replies: 1
Views: 396

Re: Atomic spectra pre assessment

To solve for the wavelength of the Krypton, you need to divide 1,650,763.73 wavelengths into 1 meter (1/1,650,763.73)- this will give you the length in meters of each wavelength, which should be 6.06x10^-7m (606nm, which is in the visible light region). Next to find the energy of one photon, you nee...
by miznaakbar
Fri Oct 13, 2017 11:24 pm
Forum: Photoelectric Effect
Topic: Difference between Photoelectric effect and Atomic Spectra [ENDORSED]
Replies: 2
Views: 418

Re: Difference between Photoelectric effect and Atomic Spectra [ENDORSED]

The atomic spectrum is essentially the range of different frequencies of EM radiation that are absorbed or emitted by electrons as they transition between energy levels in an atom. The Photoelectric Effect is simply referring to a phenomena where a metal surface will eject electrons when hit with li...
by miznaakbar
Thu Oct 05, 2017 11:37 pm
Forum: Molarity, Solutions, Dilutions
Topic: G7 - Homework Problem
Replies: 5
Views: 788

Re: G7 - Homework Problem

From what I understand, the order of mixing a solute and solvent should not matter since the amounts of each should be the same regardless of when they are put in the beaker.
by miznaakbar
Thu Oct 05, 2017 11:32 pm
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: E7 [ENDORSED]
Replies: 9
Views: 1297

Re: E7 [ENDORSED]

A TA and UA said to only round off to the number of sig figs you need at the very end of the problem, that way you can make the answer as accurate as possible. Also you're supposed to have as many sig figs in the answer as there are in the given numbers with the least sig figs. So for example, if yo...

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