Search found 41 matches

by Jeremy_Guiman2E
Sun Jan 19, 2020 6:11 pm
Forum: Ideal Gases
Topic: homework #3
Replies: 15
Views: 75

Re: homework #3

Homework 3 should probably be from the Acids and Bases Equilibria section given that we finished covering it last week (and we already completed the Chemical Equilibrium section as well).
by Jeremy_Guiman2E
Sun Jan 19, 2020 5:55 pm
Forum: *Titrations & Titration Calculations
Topic: Acid/Base bonds
Replies: 3
Views: 13

Re: Acid/Base bonds

Strong acids by definition completely (or nearly completely) dissociate in solution. Conversely, weak acids would dissociate much less. Less dissociation as with weak acids means that the molecules are still held together and would therefore have stronger, shorter bonds.
by Jeremy_Guiman2E
Sun Jan 19, 2020 5:49 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Reaction Shifts Left or Right
Replies: 7
Views: 24

Re: Reaction Shifts Left or Right

Increasing temperature adds heat to the reaction. Because the reverse reaction is endothermic, adding heat means it is inputted into the system and will favor the reverse reaction.
by Jeremy_Guiman2E
Sun Jan 19, 2020 4:49 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6B. 3
Replies: 2
Views: 27

Re: 6B. 3

This is how I did it, and your process looks good!
by Jeremy_Guiman2E
Mon Jan 13, 2020 5:47 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 5I.13
Replies: 3
Views: 51

Re: 5I.13

Thank you Brian, this was really helpful!
by Jeremy_Guiman2E
Sun Jan 12, 2020 10:10 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: stoichiometric coefficients
Replies: 5
Views: 25

Re: stoichiometric coefficients

how are stoichiometric coefficients involved in Equilibrium? Why when they are the same, its still in equilibrium? Stoichiometric coefficients are involved in the equilibrium equation as they are the exponents used for the concentrations. It is important to ensure the equation is balanced first. To ...
by Jeremy_Guiman2E
Sun Jan 12, 2020 10:02 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Calculating K
Replies: 9
Views: 20

Re: Calculating K

Pure liquids and solids don't affect the reactant amount as their concentrations stay the same.
by Jeremy_Guiman2E
Sun Jan 12, 2020 9:56 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5I.27
Replies: 8
Views: 33

Re: 5I.27

Yes, we use initial concentrations of given compounds/substances to determine their final concentrations at equilibrium. These final concentrations are part of the equilibrium composition.
by Jeremy_Guiman2E
Sun Jan 12, 2020 9:50 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: memorization
Replies: 12
Views: 37

Re: memorization

K values will most likely be given as they are unique values depending on the equations.
by Jeremy_Guiman2E
Sun Jan 12, 2020 9:41 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 5I.13
Replies: 3
Views: 51

5I.13

5I.13 (a) In an experiment, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L and heated to 1000. K to study its dissociation into Cl atoms. Use the information in Table 5G.2 to calculate the equilibrium composition of the mixture. (b) If 2.0 mmol F2 was placed into the reaction vess...
by Jeremy_Guiman2E
Tue Dec 04, 2018 12:13 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Lewis Structure for I3-
Replies: 1
Views: 662

Re: Lewis Structure for I3-

This seems to be the correct electron dot diagram: https://qph.fs.quoracdn.net/main-qimg-d14aef43ff5b2ed2d29871b3296d8571 The outer I's have a formal charge of 0, and I believe the formal charge of -1 is ok on the central I atom because of the fact that I 3 - is a negatively charged molecule, with a...
by Jeremy_Guiman2E
Tue Dec 04, 2018 12:03 am
Forum: Trends in The Periodic Table
Topic: Atomic vs ionic radius
Replies: 6
Views: 306

Re: Atomic vs ionic radius

Yes, essentially anion radius > neutral atom radius > cation radius. Atomic radius increases down a group (as amount of electrons/shells increases) and decreases from left to right (as more protons draw in electron shells more).
by Jeremy_Guiman2E
Mon Dec 03, 2018 11:59 pm
Forum: SI Units, Unit Conversions
Topic: bond angles
Replies: 5
Views: 388

Re: bond angles

This chart also sums up the point made above. The presence of lone pairs typically alters the bond angles to less/greater than expected, except in the two cases seen (steric number 5, linear shape and steric number 6, square planar shape) as these electrons can be placed anywhere on the axis, so the...
by Jeremy_Guiman2E
Mon Dec 03, 2018 4:19 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: 6th Edition 17.37
Replies: 2
Views: 81

6th Edition 17.37

17.37 With the help of Table 17.4, determine the coordination number of the metal ion in each of the following complexes:

(a) [NiCl4]2-
(b) [Ag(NH3)2]+
(c) [PtCl2(en)2]2+
(d) [Cr(edta)]-


How would you determine the coordination number for c and d?
by Jeremy_Guiman2E
Sun Dec 02, 2018 11:17 pm
Forum: Bronsted Acids & Bases
Topic: 6th ed: Fundamentals J.1
Replies: 4
Views: 123

Re: 6th ed: Fundamentals J.1

If something can produce/donate an H+ ion, it is probably a Bronsted acid. But, if it can accept one, it is a Bronsted base. The base is often OH-, but as you can seen, ammonia (NH3) can also act as a Bronsted base (proton acceptor) to produce ammonium ion, NH4+.
by Jeremy_Guiman2E
Sun Dec 02, 2018 8:25 pm
Forum: Lewis Acids & Bases
Topic: 6th editions Fundamentals J #1
Replies: 3
Views: 93

Re: 6th editions Fundamentals J #1

Recall that Bronsted acids are proton (or H + ) donors, and that Bronsted bases are proton (or H + ) acceptors. I believe that H 2 SO 3 can dissolve into sulfite ion and H + ions, while Ca(OH) 2 can dissolve into Ca 2+ and OH - . Thus, because H 2 SO 3 can donate protons, it is a Lewis acid. And, be...
by Jeremy_Guiman2E
Sun Dec 02, 2018 7:57 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Ozone and polarity
Replies: 3
Views: 196

Re: Ozone and polarity

Is the partial positive on the central O and the partial negative on the outer O just due to a greater amount of lone pairs on the outer O, or are there more/other reasons?
by Jeremy_Guiman2E
Sun Nov 25, 2018 9:50 am
Forum: Dipole Moments
Topic: canceling dipoles
Replies: 2
Views: 89

Re: canceling dipoles

Dipoles would cancel each other out when the polar bonds in a molecule are equal and opposite. Molecules that have symmetry thus often have dipole cancellation. This can be seen in CO 2 as it is a linear shape. In the CO 2 molecule, the partial negative oxygens will have a dipole moment outward, eff...
by Jeremy_Guiman2E
Tue Nov 13, 2018 10:32 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Homework 6.1 c
Replies: 1
Views: 78

Re: Homework 6.1 c

https://pubchem.ncbi.nlm.nih.gov/image/imgsrv.fcgi?cid=1089&t=l I used the above for reference. The 2 hydrogens are bonded to their respective oxygens. This is enough for hydrogen bonding because the hydrogens will get their partial positive as O is the most electronegative element present. All...
by Jeremy_Guiman2E
Tue Nov 13, 2018 10:23 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Shape
Replies: 5
Views: 112

Re: Shape

They're the same thing. You can use them interchangeably.
by Jeremy_Guiman2E
Tue Nov 13, 2018 10:00 pm
Forum: Electronegativity
Topic: Electronegativity in Molecules with N,O,F
Replies: 4
Views: 167

Electronegativity in Molecules with N,O,F

Would a molecule like nitrogen trifluoride (NF3) have an electronegativity difference (partial positive/partial negative) suitable for hydrogen bonding (i.e. with water)? I'm just not sure if the fact that these are 2 of the most electronegative elements affects this.
by Jeremy_Guiman2E
Tue Nov 13, 2018 9:07 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Repulsion Strength
Replies: 3
Views: 103

Repulsion Strength

Why does repulsion strength occur in this way: lone-lone pair > lone-bonding pair > bonding-bonding pair?
by Jeremy_Guiman2E
Tue Nov 13, 2018 7:53 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Question 4.5b Sixth Edition
Replies: 1
Views: 68

Re: Question 4.5b Sixth Edition

Yes, you'd be determining the bond angle for the ClO2+ ion.

The textbook calls it the OClO bond angle as it's just saying that the bond exists in this way (O - Cl - O), suggesting that Cl is the central atom with an O on both sides. Instead of saying angle ABC, it's saying Angle OClO.
by Jeremy_Guiman2E
Wed Nov 07, 2018 10:50 pm
Forum: Bond Lengths & Energies
Topic: Lone Pairs and Bond Strengths
Replies: 2
Views: 78

Re: Lone Pairs and Bond Strengths

Lone pairs on neighboring atoms will repel each other. This causes the bonds to weaken as the electron pairs try to move farther apart from one another.
by Jeremy_Guiman2E
Tue Oct 30, 2018 8:47 pm
Forum: Ionic & Covalent Bonds
Topic: 6th Edition, Problem 3.25
Replies: 1
Views: 73

Re: 6th Edition, Problem 3.25

The roman numerals after these elements indicate the ion being used. Because the element given has a roman numeral after it, you know it has multiple ions (i.e. Fe 2+ and Fe 3+ could be iron(II) and iron (III). For example, indium (III) sulfide will form In 2 S 3 from In 3+ and S 2- . Thus, you know...
by Jeremy_Guiman2E
Tue Oct 30, 2018 8:38 pm
Forum: Ionic & Covalent Bonds
Topic: Sixth Edition 3.11
Replies: 1
Views: 72

Re: Sixth Edition 3.11

So for a and b of this problem, we're given: (a) [Ar]3d 6 ; (b) [Ar]3d 5 (and that these are M 3+ where M is a metal). This means that to find the neutral atom (so we know which element we're looking for) we'd need to look for the electron configuration with 3 more electrons than what we were given ...
by Jeremy_Guiman2E
Tue Oct 23, 2018 8:48 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Rydberg Constant
Replies: 3
Views: 141

Re: Rydberg Constant

No problem!
by Jeremy_Guiman2E
Tue Oct 23, 2018 8:25 pm
Forum: DeBroglie Equation
Topic: HW Help 1.25
Replies: 4
Views: 184

Re: HW Help 1.25

Essentially, your set up would look like this: E = (5.00 x 10-3 g Na) (1 mol Na / 22.99 g Na) (6.022 x 1023 atoms/mol) (3.37 x 10-19 J/atom) = 44.1 J

The 3.37 x 10-19 J/atom comes from your answer in part a. Hope this helps!
by Jeremy_Guiman2E
Tue Oct 23, 2018 6:20 pm
Forum: Properties of Electrons
Topic: Speed of an Electron (hw 1.43)
Replies: 6
Views: 2233

Re: Speed of an Electron (hw 1.43)

You would just use the mass of an electron, which is m = 9.109 383 × 10−31 kg. Hope this helps!
by Jeremy_Guiman2E
Tue Oct 23, 2018 6:15 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Rydberg Constant
Replies: 3
Views: 141

Re: Rydberg Constant

The Rydberg constant is 3.29 x 10 15 when using frequency = R (1/n 1 2 - 1/n 2 2 ). Because c = wavelength x frequency, there is another form of the equation. Plugging in c/wavelength for what was frequency and rearranging the equation, we'd get: 1/wavelength = R/c x (1/n 1 2 - 1/n 2 2 ). The R/c va...
by Jeremy_Guiman2E
Tue Oct 23, 2018 6:03 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: When to double the parameter to get delta v
Replies: 3
Views: 279

Re: When to double the parameter to get delta v

Yeah, as Chem Mod said above, "uncertainty in velocity is the difference between the highest and lowest possible values for the velocity". So given 3.24 +/- 0.06 m/s, our highest possible value is 3.30 and our lowest is 3.18. The difference of the above values is 0.12 m/s which is what del...
by Jeremy_Guiman2E
Mon Oct 22, 2018 3:40 pm
Forum: Student Social/Study Group
Topic: Number 1.55
Replies: 2
Views: 160

Re: Number 1.55

For part a, the frequency = 3600cm -1 * c So, frequency = 3600cm -1 * 2.998 x 8 8 m/s = 3600cm -1 * 2.998 x 10 8 cm/s (Here, I converted everything to cm because it doesn't really matter, but you can convert to meters too.) This gives you about 1.1 x 10 14 s -1 when rounded and using correct sig fig...
by Jeremy_Guiman2E
Fri Oct 19, 2018 4:52 pm
Forum: Photoelectric Effect
Topic: help on 1.15 6th edition
Replies: 2
Views: 89

Re: help on 1.15 6th edition

They want you to find the initial and final values of "n", so you know you'd need the Rydberg equation. But, in order to use the Rydberg equation, we'd use the frequency. So, because we're given 102.6 nm, that equals 102.6 x 10 9 m. And plugging it in to: c = lambda * nu OR speed of light ...
by Jeremy_Guiman2E
Tue Oct 16, 2018 4:38 pm
Forum: Administrative Questions and Class Announcements
Topic: Test #2
Replies: 2
Views: 77

Re: Test #2

Yeah, test 2 starts Tuesday, October 23 in your discussion section. See:

https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Chem14ATestSchedule.pdf
by Jeremy_Guiman2E
Tue Oct 16, 2018 4:28 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Planck's Constant
Replies: 4
Views: 78

Re: Planck's Constant

I believe he uses 6.63 X 10-34 when there are only a maximum of three significant figures are given. Otherwise, if given four or more, he will use 6.626 X 10-34.

Regardless, I would agree though that using 6.626 X 10-34 consistently will give the most accurate results.
by Jeremy_Guiman2E
Fri Oct 12, 2018 9:55 am
Forum: SI Units, Unit Conversions
Topic: Should Significant Figures be affected by constants?
Replies: 10
Views: 275

Re: Should Significant Figures be affected by constants?

Significant figures are not affected by constants similar to how measured values like 1m = 100 cm are not used as sig figs. You would just need to look at what numbers you are given in the actual question itself as constants do not affect the sig figs.
by Jeremy_Guiman2E
Sun Oct 07, 2018 5:09 pm
Forum: SI Units, Unit Conversions
Topic: Avogadro's Number [ENDORSED]
Replies: 11
Views: 464

Re: Avogadro's Number [ENDORSED]

Yes, because Avogadro's number is used as a conversion factor, it is reversible. So, going from "x" atoms to moles, you would say "x" atoms times (1 mol/6.022 x 10 23 atoms) = y mol. Let's say you have 15 atoms. 15 atoms x (1 mol/6.022 x 10 23 atoms) = 2.5 x 10 -23 mol. Alternati...
by Jeremy_Guiman2E
Sun Oct 07, 2018 4:40 pm
Forum: Limiting Reactant Calculations
Topic: Calculating Molar Mass in Limiting Reactant Problem
Replies: 3
Views: 117

Re: Calculating Molar Mass in Limiting Reactant Problem

It looks like you are talking about the CaC 2 + 2H 2 0 --> Ca(OH) 2 + C 2 H 2 problem. In class, he calculated the molar mass of C 2 H 2 but not Ca(OH) 2 . He didn't calculate the molar mass of calcium hydroxide (Ca(OH) 2 ) because it's not relevant to answering the question. Certainly, for purposes...
by Jeremy_Guiman2E
Mon Oct 01, 2018 9:25 pm
Forum: Molarity, Solutions, Dilutions
Topic: Molarity
Replies: 3
Views: 257

Re: Molarity

Ok, so we are given 5.15 g of epsom salts, and know the formula unit would be MgSO 4 + 7H 2 O. We will use the molar mass (246.48 g/mol) to determine how many total moles of epsom salts we have. So, 5.15 g / 246.48 g/mol = 0.0209 mol epsom salts. However, because there are 7 water molecules in each ...
by Jeremy_Guiman2E
Mon Oct 01, 2018 9:08 pm
Forum: Empirical & Molecular Formulas
Topic: Empirical and Molecular Formulas Module Post-Assement
Replies: 1
Views: 415

Re: Empirical and Molecular Formulas Module Post-Assement

In this problem, we are given several mass percentages as well as the molar mass of the compound, L-Dopa, tasked to find its molecular formula. First, we will assume that we have 100 g of the substance because our mass percentages total to 100% . So, we have 54.82g C, 5.62g H, 7.10g N, and 32.46g O....
by Jeremy_Guiman2E
Mon Oct 01, 2018 8:49 pm
Forum: Balancing Chemical Reactions
Topic: Post Module Assessment
Replies: 3
Views: 227

Re: Post Module Assessment

So, we are given that there are 4 moles of butane, C 4 H 10 , as well as the chemical equation with varying stoichiometric coefficients. What we know so far, thus, is: 4 C 4 H 10 (g) + ___O 2 (g) --> ___CO 2 (g) + ___H 2 0(g) Typically, we can balance chemical equations using the CHO technique (Carb...

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