Real

If you look closely at the details in Question 14 of this packet, we are comparing the binding energy of the 1s electrons in a nitrogen atom with the binding energy of the 1s electrons in an oxygen atom. This difference in binding energy for the 1s electrons (which are located in energy level 1) can be attributed to the greater charge magnitude of the positively charged oxygen nucleus. Oxygen has one more proton than nitrogen does. In Period 2, there are two anomalies in the general trend for ionization energy values. One of these anomalies involves the elements nitrogen and oxygen. The 1st ionization energy of an atom involves the removal of the outermost electron from the atom. With nitrogen and oxygen, that outermost electron is removed from the 2p sublevel. I discuss those two anomalies in the trend for ionization energy in my video for Topic 1.7. Here is the link to that particular section of that video: kzbin.info/www/bejne/oJmacqR6h7RgiLM

@@mrfarabaugh thanks for the reply!

The chemical formula of calcium chloride dihydrate is CaCl₂•2H₂O For every one mole of this substance, there are TWO moles of H₂O 2 moles of H₂O = (2)(18.016) = 36.032 g 1 mole of this substance = (40.08)+(2)(35.45)+(36.032) = 147.012 g Mass percent of H₂O in this substance = (36.032)/(147.012)x100 = 24.51%

I just tried that link, and it seems to work okay for me: bit.ly/3z8RipZ

@@mrfarabaugh Thank you!

same whens ur sol

@@911.porschelover today and I passed

Since that question was looking for evidence that the pH of the reaction mixture is greater than 7 at the equivalence point, we need to see the fluoride ion (F⁻) behaving as a base (H⁺ acceptor) in aqueous solution. The equation (HF + OH⁻ ⇄ F⁻ + H₂O) only shows the chemical formulas of the products of the reaction that occurred in this titration experiment. However, that equation does not show evidence of the fluoride ion reacting with water as a weak base, which results in a pH > 7. That's why we need to see this: F⁻ + H₂O ⇄ HF + OH⁻ You might be thinking, "But isn't this equation just the REVERSE of the other one?" Yes, but showing that hydroxide ions are produced when fluoride ions react with water provides evidence that the pH > 7 at the equivalence point in that titration experiment. I hope that helps.

@@mrfarabaugh Thank you! I just took my AP Chem exam today. Your videos were very helpful.

You would still get full credit for that answer of -3570 kJ per mole of reaction. When a calculation involves multiple steps from the beginning to end, I try to follow a procedure in which I wait until the very end to round off my final answer. Slight variations occur naturally due to rounding numerical values. It is reasonable to expect such variations in a multi-step calculation.

@@mrfarabaugh Ok thank you!

@@mrfarabaugh I did 31.2/0.0175 to find per 1 mole and got -3560 as my answer instead of -3570. Is this my mistake or yours?

@@the_negus Those two answers are similar, but slightly different based on rounding. When we round off the answers to our intermediate values, it can cause slight variations in the final answer. Don't worry about this issue, because you definitely understand how to do the calculations. -31.2 / 0.0175 = -1783 -1783 x 2 = -3566

It's a work in progress. I will definitely create them, and upload them to my channel. I'm not sure if I will be able to complete them all before May 6th.

Some of the solid calcium hydroxide likely dissolved when additional water was added to the saturated solution. However, equilibrium has been re-established in the new solution. There is a distinction between the AMOUNT of solid (moles, grams) on the bottom of the beaker and the CONCENTRATION of the calcium hydroxide in the saturated solution. The concentration of calcium hydroxide must be the same in both beakers because both solutions are saturated and at the same temperature.

​@@mrfarabaugh Ohh I see! So though the AMOUNT of the calcium hydroxide solid does decrease, the concentration remains the same, as it takes up the same "amount/percentage" of the solution? That is, mol/L?

@@kaitlyn8185 Yes. You can think of the Ksp value (at 25°C) for an ionic compound as a property of that compound. It applies to the equilibrium system of being a saturated solution of that compound. Once you have established a saturated solution of an ionic compound, then the concentration (mol/L) of the ions in that saturated solution (at 25°C) is only dependent on the value of Ksp itself. The concentration of the ions is NOT dependent on the volume of the solution. In fact, the specific reason **that I wrote** parts (d) and (e) of Question 2 in this packet was to address that issue and the potential misconception associated with it!! Thank you for your comments.

Yes. My goal is to add short MCQ practice sets that will include the Topics in all 9 of the AP Chemistry units. It's definitely a work in progress.

Not yet. I still need to complete that packet for the Unit 9 Summative Assessment Practice and upload my video to my CED playlist. It should be available by the end of this month (April 2024).

@@mrfarabaugh awesome thanks!

kzbin.info/www/bejne/p5-9aZylbJynaJo&ab_channel=MichaelFarabaugh

According to the AP Chemistry curriculum, there are six STRONG ACIDS that students should memorize: HCl, HBr, HI, HClO₄, HNO₃, and H₂SO₄. These acids have rather large Ka values and are 100% ionized in aqueous solution. The six conjugate bases of these six strong acids (chloride, bromide, iodide, perchlorate, nitrate, and hydrogen sulfate) are so weak, that they are considered as negligible bases, and don't affect the pH of the solution. Examples of STRONG BASES are metal hydroxides in which the metal is located in Groups 1 & 2, such as NaOH, KOH, Ca(OH)₂, and Sr(OH)₂. When these ionic compounds dissolve in water, they split up to produce cations and hydroxide ions. The conjugate acid of the hydroxide ion is water. Everything else that is classified as an acid or a base tends to fall into the WEAK category. I consider acetic acid to be a weak acid (Ka = 1.8E-5) and the acetate ion (Kb = 5.6E-10) to be a weak base. I consider ammonia to be a weak base (Kb = 1.8E-5) and the ammonium ion (Ka = 5.6E-10) to be a weak acid. When I compare two different weak acids or two different weak bases by looking at their relative K values, I use words like stronger and weaker as comparative words. Acetic acid is a stronger acid than the ammonium ion (but they are both classified as weak acids). Ammonia is a stronger base than the acetate ion (but they are both classified as weak bases).

​@@mrfarabaugh Ohh I see, so I should use the Ka or Kb, and the acids and bases AP wants us to know to determine the relative strength of the acid or base? Thank you for taking the time to reply!!

Think of it like this. Some substances such as sugar or ethanol are very soluble in water, but they are nonelectrolytes. They mix with water molecules very well to form a homogeneous solution, but they don’t break up into ions. Some substances such as NaCl or KNO₃ are very soluble in water and also strong electrolytes. They mix with water molecules very well to form a homogeneous solution and they also will break up completely into positive and negative ions. Then there are weak acids and weak bases such as vinegar and ammonia.They also mix with water very well. But they ionize to a small extent. Since they are only partially ionized in solution, they are weak electrolytes. So the ability to form a homogeneous solution is one property that is unrelated to a substance’s ability to break up or ionize in the solution. With sugar or ethanol, they can form hydrogen bonds with water. With ionic substances, they can form ion-dipole forces with water. With weak acids and bases it can be a combination of these attractive forces.

In Topic 5.9, the usual situation that requires that algebraic substitution procedure involves a mechanism in which the first step is a fast equilibrium followed by a second step that is the slow rate-determining step. Although each elementary step in a proposed mechanism should have its own specific rate law and rate constant, we don’t normally get these values from experimental data. That’s because we don’t normally have the ability to carry out the elementary steps as single events. Instead, we can perform the overall reaction under various conditions in order to obtain the overall rate law. An overall rate law has a single rate constant. This number is simply a proportionality constant that is determined from our kinetics experiments for the overall reaction.