Search found 21 matches

by Michael Koger 3K
Sun Dec 10, 2017 8:29 pm
Forum: Properties & Structures of Inorganic & Organic Bases
Topic: Strong Acids [ENDORSED]
Replies: 1
Views: 250

Re: Strong Acids [ENDORSED]

A strong acid is an acid that completely dissociates in water. All of the hydrogen halides are strong acids, with the exception of HF. Nitric acid, sulfuric acid, perchloric acid, and chloric acid are also strong acids.

I hope this helps!
by Michael Koger 3K
Thu Dec 07, 2017 11:28 pm
Forum: Lewis Structures
Topic: HClO Lewis Structure
Replies: 1
Views: 225

Re: HClO Lewis Structure

Halogens are usually not the central atom because they are only capable of forming one bond (except in the case of octet rule exceptions).

I hope this helped!
by Michael Koger 3K
Thu Dec 07, 2017 11:23 pm
Forum: Hybridization
Topic: Review Worksheet Ch.3&4
Replies: 2
Views: 190

Re: Review Worksheet Ch.3&4

For every single, double, or triple bond, there is one sigma bond. A double bond also contains one pi bond, and a triple bond contains two pi bonds. When figuring out the hybridization of an atom, the first step is to count the number of electron density regions. For the first carbon atom pointed at...
by Michael Koger 3K
Sat Dec 02, 2017 9:11 pm
Forum: Amphoteric Compounds
Topic: Amphiprotic compounds
Replies: 3
Views: 334

Re: Amphiprotic compounds

The defining characteristic of amphiprotic compounds is that they can accept or donate protons depending on the chemical environment. If exposed to a strong base, it will act as an acid and donate a proton. If exposed to a strong acid, it will act as a base and accept a proton.
by Michael Koger 3K
Sat Dec 02, 2017 9:02 pm
Forum: Bronsted Acids & Bases
Topic: 12.17
Replies: 1
Views: 126

Re: 12.17

Metallic oxides are usually basic, while non-metallic oxides are usually acidic. There are several elements with intermediate electronegativities near the middle of the table that exhibit amphoteric properties. Many of these amphoteric elements are semimetals, but not all of them (e.g. aluminum). Ho...
by Michael Koger 3K
Sat Nov 25, 2017 11:58 pm
Forum: Naming
Topic: 17.31 -- Adding molecules before/after complex
Replies: 1
Views: 93

Re: 17.31 -- Adding molecules before/after complex

Since the oxidation state of the chromium atom in the complex is 3+ and the charge of each of the six cyanide ions is 1-, the overall charge of the complex is 3-. This means that three positively-charged potassium ions are required to cancel out this negative charge.
by Michael Koger 3K
Sat Nov 25, 2017 11:48 pm
Forum: Ideal Gases
Topic: Units for the formula
Replies: 4
Views: 372

Re: Units for the formula

Though these are not the SI units used for the Ideal Gas Law, they are the units most commonly used.

- Atmospheres (atm) or bars for pressure
- Liters (L) for volume
- Moles (mol) for n
- (L x atm)/(K x mol) or (L x bar)/(K x mol) for the constant
- Degrees Kelvin (K) for temperature
by Michael Koger 3K
Tue Nov 21, 2017 1:42 am
Forum: Hybridization
Topic: Question 4.81
Replies: 2
Views: 138

Re: Question 4.81

The hybridization of carbon is what they have emphasized the most in lecture and discussion, so it is more likely to be on Test 4 or the final. However, understanding how hybridization works in non-carbon atoms is also very important.
by Michael Koger 3K
Tue Nov 21, 2017 1:31 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Rate Constant Clarification
Replies: 1
Views: 92

Re: Rate Constant Clarification

To find the Kc of the reaction, you would have to convert the partial pressure values for the gases to concentration values using the Ideal Gas Law (PV=nRT).
by Michael Koger 3K
Mon Nov 13, 2017 2:27 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Radicals and electrons
Replies: 3
Views: 197

Re: Radicals and electrons

The best way to determine where the lone electron goes in a radical is to compare the number of bonds an atom is capable of forming with the number it actually forms in the molecule. If the latter number is less than the first, then it is very likely that that atom has an unpaired valence electron. ...
by Michael Koger 3K
Mon Nov 13, 2017 2:19 am
Forum: Octet Exceptions
Topic: Radical [ENDORSED]
Replies: 2
Views: 176

Re: Radical [ENDORSED]

A radical is defined as a molecule (or atom) with at least one unpaired electron. The methyl group is an example of a radical. Though the central carbon atom in the methyl group is able to form four bonds, only three hydrogen atoms bond with it, leaving it with one unpaired valence electron. Determi...
by Michael Koger 3K
Sun Nov 05, 2017 6:49 pm
Forum: Octet Exceptions
Topic: The expanded octet in PCl5
Replies: 2
Views: 311

Re: The expanded octet in PCl5

In order for the P atom to form bonds with 5 Cl atoms, it must have an expanded octet. Every one of the five bonds constitutes two electrons, so the P atom has two electrons beyond the octet. Since P is a 3rd period element, it can fill one of its empty d orbitals with these two extra electrons.
by Michael Koger 3K
Sun Nov 05, 2017 6:24 pm
Forum: Lewis Structures
Topic: Carbon as a Central atom
Replies: 1
Views: 83

Re: Carbon as a Central atom

In most cases, the central atom is the one with the lowest electronegativity. Carbon has a significantly lower electronegativity than elements such as oxygen and fluorine, so it often the central atom.
by Michael Koger 3K
Sat Oct 28, 2017 11:01 pm
Forum: Lewis Structures
Topic: Central Atom [ENDORSED]
Replies: 3
Views: 157

Re: Central Atom [ENDORSED]

The atom that requires the most electrons to fill its octet is usually the central atom. For example, carbon needs four more electrons to fill its octet, so it is often the central atom. This need for electrons is what allows carbon to form four bonds.
I hope this helped!
by Michael Koger 3K
Mon Oct 16, 2017 8:47 pm
Forum: DeBroglie Equation
Topic: How do we get E=pv? [ENDORSED]
Replies: 2
Views: 199

Re: How do we get E=pv? [ENDORSED]

This equation comes from the definition of energy. 1 joule (the unit of energy) is defined as 1 N x m, or 1 [(kg x m^2)/s^2]. Momentum times velocity also yields [(kg x m^2)/s^2]. Therefore, the equation E = pv is merely a restatement of the units of energy. The formula for kinetic energy is not use...
by Michael Koger 3K
Mon Oct 16, 2017 8:36 pm
Forum: Photoelectric Effect
Topic: Post Module #19
Replies: 3
Views: 181

Re: Post Module #19

In order to understand the question, it is necessary to understand how the photoelectric effect works. Basically, light is shone on a metal, and the metal emits electrons. However, the appropriate frequency and therefore energy of light that is needed to knock off electrons is different for every su...
by Michael Koger 3K
Fri Oct 13, 2017 1:50 am
Forum: DeBroglie Equation
Topic: DeBroglie Relation
Replies: 1
Views: 127

Re: DeBroglie Relation

I'm not sure if this will help, but here goes. de Broglie based his idea of matter waves on two equations, both of which give the energy of a photon. E = h x v E = c x p (This is used in situations involving relativity and is beyond my understanding.) 1. c x p = h x v 2. c = (h x v)/p 3. c/v = h/p 4...
by Michael Koger 3K
Fri Oct 13, 2017 1:29 am
Forum: DeBroglie Equation
Topic: Further questions about 1.33
Replies: 1
Views: 115

Re: Further questions about 1.33

a) Joule = N x m = kg x [(m^2)/(s^2)] Therefore, J x s = kg x [(m^2)/s], and (J x s)/[kg x (m/s)] = m. I think that you accidentally put J/s instead of J x s. b) de Broglie wavelength = Planck's constant/momentum de Broglie wavelength = (6.626 x 10^-34 J x s)/ [(9.11 x 10^-31 kg) x (3.6 x 10^6 m/s)]...
by Michael Koger 3K
Sat Oct 07, 2017 2:54 pm
Forum: Empirical & Molecular Formulas
Topic: Faster Method to Calculate the Molecular Formula [ENDORSED]
Replies: 2
Views: 205

Re: Faster Method to Calculate the Molecular Formula [ENDORSED]

There is indeed another way to calculate the molecular formula, but the speed of this method vs. the traditional method depends a lot on the person. This is how it goes: Step 1: By using the mass percentages and molar mass given, calculate the number of grams of each element in a mole of the substan...
by Michael Koger 3K
Tue Oct 03, 2017 6:31 pm
Forum: Limiting Reactant Calculations
Topic: Problem M.25
Replies: 2
Views: 411

Re: Problem M.25

Since the molecular formula of 2-naphthol is given in the problem, we are able to calculate what the normal percent masses of C and H in a pure sample of the substance would be (I will leave the calculations up to you). As is to be expected with a sample that is contaminated, the combustion analysis...
by Michael Koger 3K
Tue Oct 03, 2017 6:12 pm
Forum: Limiting Reactant Calculations
Topic: Problem G.25
Replies: 1
Views: 210

Re: Problem G.25

(1/2)^n represents n number of times that the molecules of X are halved (as a result of n doublings of solution volume and n subsequent extractions of 10. mL of the new solution). Every time the volume is doubled and 10. mL is taken from the new solution, the number of molecules is halved. When (1/2...

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