CHEMISTRY COMMUNITY

Madelyn Gehrich 1E

Kw = [H3O+][OH-]. So if Kw=2.1 x 10^-14, this is equal to [H3O+][OH-]. This comes from the equation 2H2O (l) --> H3O+ (aq) + OH- (aq), we take x^2 = 2.1 x 10^-14, so x=1.4 x 10^-7. You then take the -log(1.4 x 10^-7) to get a pH of 6.8. Once you have the pH you can calculate the pOH knowing that pH ...

Madelyn Gehrich 1E

A Bronsted acid is a proton donor, thus the acid in the acid-base reaction would lose its H+ and generally donate it to the water molecule in the reaction to produce a hydronium ion and some other ionic compound in the products. Once you identify the Bronsted acid and base in a reaction (which would...

Madelyn Gehrich 1E

To find the change to the value of K, you have to write the formula for the concentration. It doesn't necessarily matter what the concentrations of the products and reactants are, because we know the final value of K. To calculate this change, you use the coefficients. So if the chemical equation is...

Madelyn Gehrich 1E

^To explain that a little bit, think of adding temperature to a system. Since the forward reaction is exothermic (favoring the products), that means that the system releases heat while forming the products. With the reverse reaction being endothermic, the system absorbs heat to form the reactants ag...

Madelyn Gehrich 1E

When the values of change in molar concentration and equilibrium in molar concentration are unknown (x), the answer you will get will be in terms of x. Thus, since this variable is unknown, we set it equal to 0 and use the quadratic equation to solve for x. If you have enough information from the pr...

Madelyn Gehrich 1E

Remember that molar concentration of a pure substance (solid or liquid) does not change in a reaction, thus solids and liquids not included in K expression. So an aqueous solution should be calculated by Kc.

Madelyn Gehrich 1E

You definitely should not have to memorize the table. All of the equations and constant values that we need for an exam will be given on the front page of the test. Just make sure you know how to utilize the information in the table!

Madelyn Gehrich 1E

Essentially, you would follow the rules for a Lewis structure according to the Octet rule along with its exceptions. After you draw the molecule in a basic way, then you can evaluate where your lone pairs are to see if you need to redraw it. For example, for a seesaw shape, there would be four atoms...

Madelyn Gehrich 1E

Generally, you would put the radical on the central atom.

Madelyn Gehrich 1E

Sometimes good indicator of non-polarity can be symmetry! This is seen in naturally occurring Oxygen (O2) and Nitrogen (N2), as they are symmetrical no matter which way you look at the molecule. The same for a compound like CCl4, where the bonds look the same no matter what. These are examples of no...

Madelyn Gehrich 1E

Based on principle of resonance, all resonance structures will have the same formal charge. Thus, it should not matter which structure you draw because they will all have the same formal charge, unless the question specifically asks you to draw multiple resonance structures.

Madelyn Gehrich 1E

The best way to understand this Lewis structure is in terms of the central atoms. Carbon has the lowest ionization energy, so it will be the central atoms in the molecule. We know that Hydrogens will generally be terminal atoms, so after we put the carbon at the center we can attach the NH2 with the...

Madelyn Gehrich 1E

If you think about it, even though the 4s orbital is listed first on the periodic table, the 3d orbital still comes before it in the order of electron orbitals. Thus, when you calculate the electron configuration of the cation, you have to take it from the outmost (4s) orbital rather than the 3d orb...

Madelyn Gehrich 1E

a. Thallium (III) oxide. It indicates to you that in this compound, the thallium ion will be taking the 3+ form. Knowing that oxygen forms a 2- ion, you cross the charges to form the compound Tl2O3. b. Manganese (II) chloride. It indicates to you that in this compound, the manganese ion will be taki...

Madelyn Gehrich 1E

Yes. Any deviation from ground state is an excited state of the electron. An electron gets excited and can jump to different energy levels when n increases. When n reaches infinity, or past the maximum excitement that an electron can be in for a particular atom, the electron escapes the atom altoget...

Madelyn Gehrich 1E

Keep in mind that in the row on the periodic table containing the 4s and 4p orbitals (the fourth row), the d-block begins in this row as well and is labelled differently than the s and p orbitals. The 4th row on the table goes in the order 4s, 3d, then 4p chronologically, but since 3d is located in ...

Madelyn Gehrich 1E

Think about it as if you are finding the energy of one photon in part a, and then in part b you are finding the energy of multiple photons. So since you found the energy of one photon in the first part, all you would have to do is convert 5.00 mg of sodium into number of photons, and then multiply i...

Madelyn Gehrich 1E

This would mean that the kinetic energy is equal to 0. Kinetic energy is something that is considered excess in the energy equation involving the photon, and if the kinetic energy is equal to 0 then that would be the maximum wavelength of that photon as well.

Madelyn Gehrich 1E

Are we supposed to do the Chapter 2 questions for homework this week or can it be from Chapter 1 and 2?

Madelyn Gehrich 1E

Atoms with more than one electron can emit more than one photon because they have multiple energy levels as the atomic number gets higher (thus the electron number gets higher). Thus, an atom can emit multiple photons because there are more energy levels of which the electrons can move in and out!

Madelyn Gehrich 1E

I think instead of conceptualizing it with everyday objects (because those do not move nearly as fast and are not nearly as small), it is useful to try and picture it using the subatomic particles. Say that you are trying to analyze an electron under a microscope (not an easy task). When you are try...

Madelyn Gehrich 1E

You switched the last part of the equation. You have 42.51 mol O2 but you have to multiply that by ((32.00 g O2)/(1 mol O2)). In your equation you did division instead of multiplication for the grams of oxygen gas. If you fix this, it should give you 1.33 x 10^3 g O2!

Madelyn Gehrich 1E

You would use 3 sig figs. You always use the least amount of sig figs from the figures involved in your calculations for your final answer!

CHEMISTRY COMMUNITY

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Re: 12.23 [ENDORSED]

Re: Expanding Equations [ENDORSED]

Re: 11.37

Re: Le Chatelier's Principle

Re: Equilibrium Composition

Re: Aqueous in Kp?

Re: Homework 11.25 [ENDORSED]

Re: Lone Pairs

Re: Radicals

Re: 4.27

Re: Resonance Structures for the Midterm

Re: Problem 2.37c Lewis Structure for H2C(NH2)COOH

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Re: What exactly is an excited state? [ENDORSED]

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Homework Week of 10/16 [ENDORSED]

Re: Calculations with non H atoms

Re: Visualizing Heisenberg Uncertainty Equation

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