CHEMISTRY COMMUNITY

Sarah_Wilen

Kw= (OH-)(H3O+). To find the OH- and H3O+ concentration, you must square root the Kw value to obtain Ka and Kb. Why is that? Because in neutral water, the concentration of H3O+ and OH- is the same. Kw=x^2 square root kw to get the value of x which is your H3O+ and OH- concentrations. Next, to find t...

Sarah_Wilen

This really clarified the salts for me. Thank you so much for sharing this, JD! :)

Sarah_Wilen

I wrote the dissociation equilibrium equation for CH3COOH, then drew an ICE table for it. I looked up the Ka online and plugged it into the dissociation constant equations. Next, I solved for x and found the value for [H3O+]. Using the now known [H3O+] value, I was able to calculate the pH and pOH.

Sarah_Wilen

A strong acid means an acid that is able to 100% ionize into its ions. Ka is the acidity constant, which is the ratio of the concentration of the reactants to the concentration of the products to the power of their stoichiometric coefficient. If the Ka constant is on the higher size, we know that th...

Sarah_Wilen

Partial pressure is the pressure that would be exerted by one of the gases in a mixture of gases by itself. For example, our atmosphere is composed of a mixture of gases. To name some there is nitrogen and oxygen gas. I don't know the exact number, but for an example, let's say we trapped atmosphere...

Sarah_Wilen

Just so you know, we didn't need to solve this problem. It wasn't included on the assignment sheet. Do you still want to know?

Sarah_Wilen

Imagine you have an enclosed flask of water at room temperature. Particles at the surface of the water in the flask are escaping from its liquid state and becoming a gas. Because the reaction is happening in an enclosed flask, the gas particles cannot float out of the flask. Once there are more part...

Sarah_Wilen

Let's dissect the question. It is actually a lot simpler than it seems. All the wording fluff makes the problem confusing. The question gives you two different systems of the same equilibrium reactions. The only thing differing is the concentration of O3. System #1 2 O_{3(g)}\rightleftharpoo...

Sarah_Wilen

Not sure if you did anything wrong, but here is my work, which matches the answer.

Sarah_Wilen

A way to quantitatively show the relationship between the strengths of conjugate acid-base pairs is like this... We know that [H_{3}O^{+}][OH^{-}=K_{w}] So it is only right that if the concentration of the acid is very high (strong acid), then the concentration of the base must be lower to make up f...

Sarah_Wilen

Oops, mistake. You are right. I didn't realize what you were asking. Yes, relatively speaking, strong acids have weak conjugate bases. Strong bases have weak conjugate acids. For example, HCl is a strong acid. Its conjugate pair is Cl-, which is a weak base. HCl(g) + H2O(l) <-----> H3O+(aq) + Cl-(aq...

Sarah_Wilen

I don't know if they will be given or not, but I made this square in high school chem. I think it helped me.

Sarah_Wilen

The word strong in front of acids and bases means that the compounds dissociate COMPLETELY into their ions. On the other hand, weak acids and bases do not completely dissociate, so the dissociation reaction is reversible. I think you are asking about conjugate acids and bases. If there is a strong a...

Sarah_Wilen

The value of Kw for water at body temperature (37°C) is 2.1 X 10^-14. (a)What is the molarity of H3O+ ions and the pH of neutral water at 37°C? (b) What is the molarity of OH- in neutral water at 37"C? How would I approach part (a) and (b)? Notice that the value of Kw shifted from 1.0 x 10^-14 ...

Sarah_Wilen

The definition of amphoteric is a species that is able to act as both a base and acid, which means it should be able to accept and donate a proton when need be. Well, for one example, we know that water is amphoteric... amphoteric2-1.png Examples: H2O can become H3O+ or OH-. HCO3- can become CO3^2- ...

Sarah_Wilen

For a chemical equilibrium reaction, when you change the coefficients of the reactants and products, you change the K constant to the power of the change. In problem c, equation #4 = 2 x equation #1. Therefore, the k constant is: K_{eq 3}=K_{eq 1}^{2} Now, you take the equilibrium constant and set i...

Sarah_Wilen

I don't think the problem will ever explicitly say that you are dealing with a strong or weak acid or base, but if you know which acids and bases are strong, then through the process of elimination, you will be able to tell if the acid or base you are given is strong or weak. Here are the strong aci...

Sarah_Wilen

Remember, what is a Bronsted-Lowry acid and base? It is the proton theory transfer fun stuff. According to the bronsted definition, an acid is a species that can transfer a proton. A species that can accept a proton is a base. The proton transfer is the H+ transfer. Now, we assess the equations... a...

Sarah_Wilen

A conjugate pair of molecules refers to two molecules that have identical molecular formulas except that one of them has an additional H+. Here are some examples: HCl and Cl-, H2O and OH-, H2PO4- and HPO42-, Na+ and NaOH. The member of a conjugate pair with an extra H+ is called the conjugate acid a...

Sarah_Wilen

Remember that the only phase species that are in the equilibrium equation would be aqueous and gaseous substances. Solids and liquids don't participate in the equilibrium equation because they don't affect it. Pure solids and liquids are disregarded and their value is kept as 1 as to not affect the ...

Sarah_Wilen

Equilibrium is when the rate of the forward reaction matches the rate of the reverse reaction. From looking at the flasks, you should know that you have reached equilibrium when it seems as though the reactions have stopped and no product or reactant is being made (even though you know because of dy...

Sarah_Wilen

Exactly, if you were to have 2 ADP and 3 P produced, you would multiply the # for ADP by 2 and P by 3. Here's an example, see if you can try it: A mixture consisting initially of 3.00 moles NH3, 2.00 moles of N2, and 5.00 moles of H2, in a 5.00 L container was heated to 900 K, and allowed to reach e...

Sarah_Wilen

keep em coming

Sarah_Wilen

The electron configuration for Ni is [Ar]3d^{2}4s^{2} . To obtain the electron configuration for Ni+2, you take off the last two electrons you see written in the e- configuration. In this case, the e- are taken off of the 4s orbital. You are left with: [Ar]3d^{8} as the e- configuration for Ni+2. Re...

Sarah_Wilen

Hey, You are on the right track in your thinking, but I think you may have read the question wrong. I have attached the question below. It is asking In+ rather than 2+. That is where you got a different answer. I'll walk through them In: [Kr]4d^{10}5s^{2}5p^{1} In^{+}: [Kr]4d^{10}5s^{2} -Needs 49 e-...

Sarah_Wilen

To my understanding, I believe that any elements in n=3 or beyond can have expanded octets. In expanded octets, the central atom can have ten e-, even twelve. Molecules with expanded octets involve a nonmetal central atom found in the third period or beyond. While the octet rule is based upon s and ...

Sarah_Wilen

There are two parts to this question: a) energy per photon b) # photons generated To find the energy of the photon, I used this equation: E_{photon}=\frac{hc}{\lambda } As for the values... -Used Planck's constant for h -Used speed of light constant for c -Converted the given wavelength of 1850 nm i...

Sarah_Wilen

First, we shall think: What is polarizability? Is it a man, a myth, a legend? Polarizability is the ease of distortion of the electron cloud of an atom or molecule by an electric field. Neutral nonpolar individuals have symmetric arrangements of electrons in the electron clouds. When the nonpolar in...

Sarah_Wilen

Are you sure you are plugging the right exponents (right number and sign?). Maybe you accidentally put positive rather than a negative sign in front of the power? I am trying it right now, and I am still getting the right answer. Maybe your calculator is haunted? Lol did you actually post this at 4 ...

Sarah_Wilen

This is because the work function's energy is given in kJ/mol. You need to solve for the energy of a single unit. That was very poorly explained, sorry. Refer to this other answer:
viewtopic.php?f=14&t=21193

Sarah_Wilen

I liked the example Dr. Lavelle used in class for shielding, so I will retell it in my own jumbled up way. He explained, imagine you are at a campfire. There is a fire pit in front of you and you are feeling this nice heat radiating from it. You like the heat and feel an inclination towards it like ...

Sarah_Wilen

If this is a question about spectroscopy for a certain element, such as hydrogen, I believe the answer would be C. Each element has a specific spectral "footprint", therefore, will only absorb or emit specific wavelengths that correspond to them. When you observe the spectral "chart&q...

Sarah_Wilen

There isn't a formula for part b and c. Instead, the energy you use the energy you calculated in part a to see how much is created using 5.00 mg of sodium (part b) and 1.00 mol of sodium (part c). In part b, you need to convert the 5.00 mg of sodium atoms into the number of Na atoms, then you can us...

Sarah_Wilen

Hey, JD, I personally like to write the answer down in meters, then convert it to nanometers, so it looks like a readable number to me. I think it may depend on the question. If the question asks for the answer in meters, then meters they shall receive. I converted the wavelength into nanometers for...

Sarah_Wilen

According to De Broglie, all matter has wavelike properties, but it is only noticed for moving objects with a very small mass like an electron. When solving problems such as the one on page 15. To determine if it has any measurable wavelike properties, you would look at the power. 10^-38 is a really...

Sarah_Wilen

You know the equation for the energy produced by the electron electron "path" is E=\frac{-hR}{n_{f}^{2}}+\frac{hR}{n_{i}^{2}} You also know that E=h\nu We can use these two equations to solve for the final energy level. The problem tells you that the excited hydrogen atom emitted light, th...

Sarah_Wilen

The last line gave you the wavelength of 2.5 nm. First, convert nanometers to meters by dividing by 10^9 nanometers. An approach would be to use the speed of light equation to solve for velocity: c=\lambda \nu \rightarrow \nu =\frac{c}{\lambda } Then, to solve for energy, use E=\frac{hc}{\lambda } ....

Sarah_Wilen

Great tips thank you! Are you allowed to write in pencil?

Sarah_Wilen

Great test taking and studying tips. Thank you! What kind of threw me off for this test was using pen. When I made a mistake, I would have to cross it out and my final answers would end up confusing and messy. Anyone have tips to counter this?

Sarah_Wilen

Like Dabin explained, the ionization energy is the amount of energy it takes to remove the outermost electron from a neutral atom. It decreases a group because the attraction between the valence electrons and nucleus is low due to the increased shielding from the core electrons (or inner electrons)....

Sarah_Wilen

The ground state in an atom is when electrons are in the lowest possible energy level. In this state, the electrons have the lowest potential energy. What I mean by low potential energy is that there is nowhere for the electron to "fall". Potential energy is the energy possessed by the ele...

Sarah_Wilen

As Paul said, speed is a constant, therefore, cannot change. If the frequency of EMR decreases, then the wavelength would increase. In the speed of light equation, the frequency is inversely proportional to wavelength. That means as one value decreases, the other increases. If wavelength decreases, ...

Sarah_Wilen

Absorb:

Sarah_Wilen

The question is: The molar mass of the metal hydroxide M(OH)2 is 74.10 g/mol. What is the molar mass of the sulfide of this metal? What you want to think is: 1) What is the question asking? -What is the metal in the metal hydroxide? -What is the metal as a "metal sulfide"? -What is the mol...

Sarah_Wilen

OOHH ok I get what you're asking. I'm too in the 4th of July mood. I used C6H9Cl3 molecule rather than AgNO3 because in order to find the mass of the product, you have to use base the calculations off of the limiting reactant. The limiting reactant dictates when the reaction stops when it is used up...

Sarah_Wilen

Hey, Daniela, I compared AgNO3 to C6H9Cl3 because you compare the reactant to the reactant to find the limiting reactant. The mass of the reactants was given, so I compared my calculated mass to the actual mass to tell which reactant was the limiting reactant. You never compare the reactant to the p...

Sarah_Wilen

To determine what fraction you need, what I would do is count up the oxygens (I need 19). 19 doesn't divide evenly, which is a problem because we have O2. So, I will do 19/2 because I know I need 19 Oxygens and I will "split" up the O2 to get 19 of them. In the end, I just multiply everyth...

Sarah_Wilen

Here are your two equations to combine: E=h\nu c=\lambda \nu You set the speed of light equation to equal velocity: c=\lambda \nu \rightarrow \frac{c}{\lambda }=\frac{\lambda \nu }{\lambda }\rightarrow \frac{c}{\lambda }=\nu Then, you see that E=h\nu has velocity in it, so you can plug the velocity ...

Sarah_Wilen

Is there a good way to see which is the limiting reagent because I'm still a bit confused on how to compare. I understand how to do the calculations until I get to the part where I compare the calculated moles with the moles from the balanced equation. As I am a bit confused can someone clarify how...

Sarah_Wilen

The change in energy is calculated as the final energy minus the initial energy. Of course, because we started at a higher principle quantum level, the initial energy was larger than the energy of the final principle quantum level. So, there was a minus sign in the change of energy to represent the ...

Sarah_Wilen

He was just explaining the unit conversion factors: There are 10^-12 meters in one picometer OR 1 meter in 10^12 picometers There are 10^-9 meters in one nanometer OR 1 meter in 10^9 nanometers There are 10^-3 meters in one millimeter OR 1 meter in 10^3 millimeters. The purpose of him telling use th...

Sarah_Wilen

Ohhhh, so the limiting reactant does not mean the number that is the least in amount rather the number that controls how much will be made? Exactly! The smaller of the initial amount of masses of reactants given may not be the limiting reactant. Sometimes, questions will give you the same amount of...

Sarah_Wilen

With the use of the fluoride toothpaste, the mass of the enamel increases when changed from fluorapatite to hydroxyapatite. The question is asking you what is the percent change of mass from hydroxyapatite to fluorapatite in order to see the increase of the mass of the enamel. The percent change equ...

Sarah_Wilen

Notes to take into account before starting: -Molarity is in moles/liter (don't forget to convert mL to liter) -1000 mg/g -1000 mmol/L We already made the solution, so the molarity of the solution won't change. We are just measuring out different amounts of solution based on what the questions asks f...

Sarah_Wilen

Molarity is measured in moles per liter. Yes, you would convert mL to liters (1000 mL per liter). :)

Sarah_Wilen

Friend, 'tis I. Let us walkthrough some potential blunders whilst solving this problem: 1) Is it balanced? I think yes 2) Did you convert CaCO3 into kg? Now I shall walkthrough some steps: 1) Balanced ok yay 2) Convert 1.00 kg of CaCO3 into grams (1.00 kg=1000 g) 3) Convert CaCO3 into moles by using...

Sarah_Wilen

Hey, friend, Let's troubleshoot together: 1) Maybe you didn't convert 1.000 kg of AgNO3 to grams and thought that it was the limiting reactant? 2) Chemistry is hard 3) Maybe you got the wrong limiting reactant? 5) Is that the equation given (meaning not balanced yet)? Or is that the equation you bal...

Sarah_Wilen

You are exactly right! I didn't catch that thank you so much. Yes, divide 1.506*10^5 by 6.02*10^23 to get the function for one electron per atom. Then, use that number instead of whatever I wrote down to get the final answer to solve for frequency. Sorry for the confusion! Thanks Since the work func...

Sarah_Wilen

Lavoisier would be shaking in his French grave if somehow products could gain magical mass. According to the Law of Conservation of Mass, mass in an isolated system is neither created nor destroyed in a chemical reaction. It wouldn't make sense for extra mass to suddenly appear in your product. For ...

Sarah_Wilen

Look at my attached image for a worked out solution. I will show you the dance step by step: 1) Balance--yay 2) Choose a quantity (either NH3 or O2). You can choose whichever you want because this is America 3) I decided to go with ammonia, but you can choose oxygen if your heart so desires 4) I use...

Sarah_Wilen

You won't have trouble soon. Limiting reactant problems have an acquired taste, but once the taste it acquired, it is delicious. I'm going to walk through the steps. 1) Identify reactants and products 2) Balance the reaction 3) You are always given two quantities for the reactants. Whether is be the...

Sarah_Wilen

Hope the image of the worked out problem attached helps. I'll walk through my brain jumble: 1) To solve for the KE of the ejected electron, use the equation on the right side of the diagram because this is the amount of KE of the electrons. This equation is E E_{KE}=\frac{1}{2}m_{electron}\times v_{...

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