CHEMISTRY COMMUNITY

Mashkinadze_1D

The pKa is the -log of the Ka value. The larger the Ka the stronger an acid, and therefore the lower its pKa.

Mashkinadze_1D

The cis form means that idealistically atoms are on one side of the compound when it is formed and the trans form menas the identical atoms are facing away from each other or on opposite sides.

Mashkinadze_1D

The maximum I can imagine we would be expected to calculate is the Ka from the Kb, in which case you simply divide Kw by Kb or 1 x 10^-14 by Kb. Just know that the larger the Ka values the higher the acidity and the that higher the Ka the lower the pKa.

Mashkinadze_1D

I believe for something like this we would need the pKb value but we can also use a rule of thumb that most compounds with N are going to be a weak base and also due to the negative charge on hypobromite we can figure it will be a stronger acid as this allows it readily accept a proton.

Mashkinadze_1D

I think you should definitely memorize the strong acids, HBr, HI, HCl, HClO4, H2SO4, and HNO3. Also it is a good rule of thumb that carbon is usually correlating to a weak acid and nitrogen to a weak base. Hope this helps a little!

Mashkinadze_1D

If you recall, Dr. Lavelle stated that the strength of an acid is determined by two aspects the first being how easily it gives off its H+ and the second being the stability of the remaining anion. In the case of HF, the anion would not be as stable as HCl because Cl- is able to withdraw much more e...

Mashkinadze_1D

Lewis acids are more encompassing term while Bronsted is more specific. Bronsted acids are those that donate an H+ or proton in a reaction, while lewis acids are electron acceptors in a reaction. Therefore, every Bronsted acid is a lewis acid but some lewis acids such as Al2O3 are not necessarily Br...

Mashkinadze_1D

If you are referring to the right tables, in the solutions book I kept getting these wrong too. I believe there may be an error for when I multiplied their concentrations I did not get 1 x 10^-14 and the product of [H30+] times [OH-] should be this.

Mashkinadze_1D

According to the ligand naming worksheet on the website, either will be accepted but generally Dr. Lavelle prefers we use isocyano.

Mashkinadze_1D

A ligand is polydentate if when we draw the lewis structure there is more than one bonding site in the form of lone pairs present on separate atoms in the compound. Fro the textbook, the main ones we have to know are en which is bidentate, bien which is tridentate, oxalate which is bidentate, and ed...

Mashkinadze_1D

The ligand that is alphabetically first is the one that we shall use first. For example ammine would go before carbonyl and sulfato. Hope this helps!

Mashkinadze_1D

A bronsted acid is a H+ donator, the bronted base is an H+ acceptor.

Mashkinadze_1D

Every single type of bond has one sigma bond, double bonds have one sigma and one pi, triple bonds have one sigma and two pi bonds. I hope this helps!

Mashkinadze_1D

The first oxide is going to be basic because it contains a group two metalic oxide. The second is going to be acidic because S is a nonmetal and therefore this forms an acidic oxide. The next two are both amphoretic because they are located near the staircase on the periodic table and can act as bot...

Mashkinadze_1D

The amphoretic compounds are going to be the ones near the staircase on the periodic table. Although it is easier if you have them memorized, if you have a periodic table we can still figure out which ones are most likely to be amphoretic. Hope this helps!

Mashkinadze_1D

Strong bases will contain column 1/2 elements. The strong acids are hydrochloric, hydrobromic, perchloric, nitric, sulfuric, and hyrodioic. For acids, you should just memorize them as there is only six and it will make it easier.

Mashkinadze_1D

Yes, I believe we should know the general structure of how the compound would look. In lecture, Dr. Lavelle emphasized the square planar, octahedral, and tetrahedral geometries and therefore I believe we could be tested on it.

Mashkinadze_1D

The donation and acception of lone pairs is essentially the movement of hydrogen. When hydrogen is moved across, the lewis acid gives off a proton and is therefore accepting a lone pair as it becomes negative. The lewis base that has now recieved a H+ is going to give off a lone pair as it becomes p...

Mashkinadze_1D

Yes, when I asked him in office hours he stated that we must know these. We should know how many times each can bond too. Hope this helps!

Mashkinadze_1D

NaCl has a higher boiling point because there is more ionic forces present due to a higher electronegativity difference. The difference between Na and Cl in electronegativity is larger than that of H and Cl making the NaCl molecule be significantly more ionic. This ionic force then causes there to b...

Mashkinadze_1D

There is definitely a connection, the larger the intermolecular forces between two molecules of the same compound the higher the boiling point will be. This is because the forces between them are stronger and therefore more kinetic energy must be added to break these bonds. Any molecule with a highe...

Mashkinadze_1D

If a molecule has a molecular geometry of linear and both of the atoms on the sides of the central atom are the same, then it will not be polar. This is because all of the dipole moments will cancel out with the vectors and therefore too this one is nonpolar. Hope this helps!

Mashkinadze_1D

The molecule is polar because of the lone pair and trigonal pyramidal molecular geometry. The dipole moments are not cancelled out because there is not one countering the two on the top, where there is the lone pair. In general if the structure is not linear and has a lone pair, it is going to be po...

Mashkinadze_1D

Yes, when I asked Doctor Lavelle at office hours he said that it is okay to simply states less than a certain angle such as less than 109.5. We are not expected to know every single angle only the general trends that they follow because this would change with each element in theory.

Mashkinadze_1D

The method through which they are bonded in the s or p suborbitals gives this characteristic. When we look at a sigma bond, we see that it does not have to be fixed in place as it is to the side of the atom, while the pi bonds are above and below a atom and therefore cannot move or they will break. ...

Mashkinadze_1D

This would be based primarily off memorization and then identification. The way we can see this is induced-dipole-induced-dipole in every aspect of every molecule due to the slight polarization in electron cloud. The next one we can look at is dipole-dipole which occurs any time that there is a sign...

Mashkinadze_1D

The induced-dipole-induced-dipole forces in the molecule with iodine are so much higher than the one with fluorine that it cancels out the stronger dipole due to the large difference in the amount of electrons between the two elements. In certain cases a stronger dipole moment would create stronger ...

Mashkinadze_1D

A trigonal planar shape has no lone pairs attached to the central atom and a T shaped has two lone pairs attached to the central atom. Similarly, a tetrahedral structure has no lone pairs attached to the central atom, while a seesaw would have one lone pair attached to the central atom. Hope this he...

Mashkinadze_1D

The polarity of the molecule refers to the manner in which the dipoles act on the appropriate structure. Unequal dipole moments would make this a polar molecule, it does not have to do with the forces between the atoms in the molecule. Therefore, yes it is possible that a nonpolar molecule can have ...

Mashkinadze_1D

For trigonal planar, if all of the atoms attached to the central atom are the same then the bond angle will be 120. A smaller atom on one of the bonds would make the angles smaller and a larger atom than the rest would make them larger. Hope this helps!

Mashkinadze_1D

This question was asked to Doctor Lavelle today during office hours. He stated that because the difference in size and therefore electron cloud between fluorine and iodine is so large, that it over-arches the difference in electromagnetically. This is a good example of just how strong induced-dipole...

Mashkinadze_1D

I believe you are referring to inter molecular forces, as dispersion forces are simply a type of IM forces. These will be gone over, but the general methodology for determining whether or not a molecule will be polar if the dipole moments in it do not cancel in terms of their vectors and a general r...

Mashkinadze_1D

If a lone pair is present then it will decrease the angles because it causes more repulsion. However, if it is a smaller atom that is present for example change F to H, then the bond angles will decrease. In contrast, if it is larger like a H to F then they will decrease the angles just as in the lo...

Mashkinadze_1D

For identifying, we begin with drawing the Lewis structure then count the amount of bonds that each atom has keeping in mind that one single bond is a sigma bond, a double bond is a sigma and pi bond, and a triple bond is a sigma and two pi bonds. From there it is purely identifying and adding. Hope...

Mashkinadze_1D

We don't particularly calculate dipole moments, but they occur primarily when there is two atoms with differing electronegativities bonded together. I can develop more if necessary, hope this helps!

Mashkinadze_1D

No, but all of the formal must add up to the overall charge of the atom. Hope this helps!

Mashkinadze_1D

In general, the first four elements along with boron's group are an exception and can have six valence electrons. Anything past, P in the periods can have more than an octet as they begin to occupy the d orbital.

Mashkinadze_1D

You would then divide by avogadro's number because 1 mol = 6.022 x 10^23 and in this case with dimensional analysis it would allow the moles to cancel. Hope this helps!

Mashkinadze_1D

In general dont forget that we draw the structure around the least electronegative element. In this case it is carbon and therefore for this reason it will be the structure we draw around. Generally, carbon likes to form four bonds and the other three elements will allow it to bond this way with a d...

Mashkinadze_1D

The number of valence electrons is all of the electrons that are not a part of the inner core. Therefore any electron past the abbreviation we used with [X] where X is a noble gas, will be a valence electron regardless of its subshell or energy level for our purposes in calculating the amount of val...

Mashkinadze_1D

I believe that in most cases, just know that the second ionization energy will always be greater than the first due to the higher nuclear effective charge acting. Otherwise, just know the general trends and what it means. Hope this helps!

Mashkinadze_1D

In general, for exceptions we can imagine what is necessary for the full octet, and for some why they can have more than an octet. For those with an incomplete octet, they are in the boron group and the reason they can form only three bonds is this benefits them as they have three valence electrons ...

Mashkinadze_1D

The electrons never "leave". This is why they are shared, the two cross paths and then enter the orbitals of the other element but to say that they were leaving would be incorrect. This is why when a bond is covalent on a Lewis structure we are able to add one of the electrons into the she...

Mashkinadze_1D

If I am correct, hydrogen generally tries to bond to N, O, or F as this allows it to take place in hydrogen bonding. Therefore, it did not bond to Cl because then the overall strength of the molecule would be decreased significantly.

Mashkinadze_1D

The way you can see this is because oxygen is going to be the fourth electron in the p orbital, it will become less geometric. The two most favorable states for the orbitals to be occupied as is half full or full. Nitrogen is half full and therefore it will be harder to take away an electron because...

Mashkinadze_1D

This would be correct. To the best of my understanding, valence electrons are those that when we draw are a part of an energy level that has not been fully occupied. Therefore, in the one where the d is already full they do not count for valence electrons but because in the other the d is not comple...

Mashkinadze_1D

We should calculate the formal charge on any compound that has a resonance structure. Through this we are able to determine which of the multiple structures will be the most stable and therefore the most favorable for us. The structure that has the least amount of charged atoms and the lesser streng...

Mashkinadze_1D

To the second question, in general, the ionization energy will decrease as we go down the periodic table due to a greater effect of shielding and therefore less done by the nucleus to keep the electrons bound. The ionization energy will increase as we go across the rows of the periodic table from le...

Mashkinadze_1D

This is due to the face that the effective nuclear charge will now play a much higher effect on the electrons. The electrons will have less shielding and therefore the connection to the nucleus will be greater and therefore we would need to add more energy to remove an electron than when shielding w...

Mashkinadze_1D

I believe that if the question asks for a structure we should draw the most stable one which can be found by calculating the formal charges. Furthermore, I believe if the question is asking for a resonance then it will be indicated.

Mashkinadze_1D

I believe that if the question asks for a structure we should draw the most stable one which can be found by calculating the formal charges. Furthermore, I believe if the question is asking for a resonance then it will be indicated.

Mashkinadze_1D

Hello it would be as follows, feel free to ask for an explanation in any case:
a) [Kr]4d10 5s1
b) [He] 2s2
c) [Kr] 4d10 5s2 5p3
d) [Ar] 3d10 4s2 4p1
e) [Xe] 4f14 5d5 6s1
f) [Kr] 4d10 5s2 5p5

Mashkinadze_1D

This is not the ground state configuration because all of the arrows should be facing up and positive in the ground state. The middle arrow looking down is the arrow that does not configure to this rule and is therefore showing a non-ground state configuration.

Mashkinadze_1D

Hello, so if we were to google the "electron configuration" there is a chart that we go by with arrows, but this is the more tedious and often confusing method to understand this. Generally the best way to understand this is all elements in groups 1-2 are s, all elements in groups 13-18 ar...

Mashkinadze_1D

These all correlate to the quantum number of l. This is simply how they are shown on the periodic table. The connection between the two is seen!

Mashkinadze_1D

Basing off the previous reply, it would be very difficult to have a symmetric orientation when we are one away from having a much more stable (5 or 10) configuration for the d block. Therefore, because of the increased stability one electron will enter the d block from the s block to make it more st...

Mashkinadze_1D

The orbital angular momentum quantum number corresponds to the subshell. For example, the s is zero, p is one, d is two, and f is three. These are representations of how the electrons are behaving around the nucleus in the electron field.

Mashkinadze_1D

The arrows will fit the amount of orbitals and each represents an electron. For example, if we potentially have five electrons in the d subshell then we would have five arrows pointing up. Due to the Pauli Exclusion Principle, all of the arrows will face up and spin positively until we have enough t...

Mashkinadze_1D

Frequency is the speed at which light travels, while intensity is the amount of light or photons present. bFor example, when we change the wavelength then we change the frequency in aspects such as different colors. In contrast, the intensity would be compared to having one light source shining ligh...

Mashkinadze_1D

The unit eV is equal to 1.602 x 10^-19 J. Use standard conversion rates and mechanics from there.

Mashkinadze_1D

Remember that: ƛ = h/p or ƛ = h/mv For this solve first for the proton wavelength ƛ = h/(1.6 x 10^-27kg)(2.75 x 10^5m/s) ƛ = 1.4 x 10^-9 Solve for the neutron wavelength ƛ = h/(1.76 x 10^-27 kg) (2.75 x 10^5m/s) ƛ = 1.4 x 10^-12 Then: (1.4x10^-9)/(1.4x10^-12) = 100% If I did this correct than this w...

Mashkinadze_1D

The way that I would approach this problem is to replace the energy with hc/ƛ which is equivalent to the energy. From there I would suggest you use the -hR/n^2 and set it up as hc/ƛ=-hR(1/1 - 1/n^2). I tried to set up the basic idea here, but if necessary I can also expand on the calculations portio...

Mashkinadze_1D

Although I may be incorrect, I believe what is wrong with your answer is the -6.17 x 10^14 as a frequency cannot be negative. If I am incorrect, I hope someone else can show the correct answer so I can have a better understanding too, but if I am, hope this helps!

Mashkinadze_1D

If I am correct, the reasoning behind this is that the value can only be over or below it. The value could never be both larger and smaller simultaneously. Therefore the uncertainty or indeterminacy would only take in the values that we are unsure of. This may be incorrect as I was slightly confused...

Mashkinadze_1D

This would be the uncertainty, because at any given time, the position of the electron can be anywhere to what it is "confined" to. For example, similarly, if we imagine the uncertainty of a ball bouncing around the box, the only positions that the ball could have would all have to be with...

Mashkinadze_1D

Yes, light is a large amount of photons traveling at the same frequency and wavelength. These photons can sometimes behave as a wave, and sometimes they can behave as a particle. Hope this helps!

Mashkinadze_1D

Correct, the electron has mass and therefore you would use De Broglie's equation. Electrons unlike photons have mass and this is listed in the textbook.Hope this helps!

Mashkinadze_1D

Under certain conditions, light will also act as a particle with photons. This was seen in the photoelectric effect when the light particles were able to remove electrons from the metal surface. Prior to that experiment by Einstein, it was largely believed that light was only a wave. Hope this helps!

Mashkinadze_1D

Similar to any other unit, you would convert to the base unit and then continue to use conversion ratios from there. For example, we would convert nanometers to meters and then the meters to Angstroms. Hope this helps!

Mashkinadze_1D

These are energy levels at which the electron is stable around the nucleus. In other areas, the electron field would be unstable. These stable levels are represented as n=1, n=2, and on.

Mashkinadze_1D

We would use the Debroglie Equation for anything that has mass. Because an electron has a mass we would use the Debroglie. In contrast, photons are practically massless objects and therefore we do not have to worry about representing their mass and can use c= ƛv. Hope this helps!

Mashkinadze_1D

Hello! I asked Doctor Lavelle this at his office hours and he essentially explained that although there are not many practical applications, it helps us understand greatly both what a photon is, how it behaves, and the overall structure of atoms. He did mention that it can help us distinguish differ...

Mashkinadze_1D

a) ƛ = h/p ƛ = h/mv 3.6 X 10^3 km x 1000m/1km = 3.6 X 10^3 m (6.62 X 10^/34 J/s)/ (9.2109x10^-31 kg) (3.6 x 10^6 m/s) = 2.0 X 10^-10 m ƛ = 2.0 X 10^-10 m b) E=hv Hz = 1/s E = (6.6 x 10^-34 J/s) (25 X 10^6 Hz) E = 1.66 X 10^-17 J c) E = 1/2 m v^2 E = 1/2 (9.10938 X 10^-31 kg) (3.6 x 10^6 m/s)^2 E = 2...

Mashkinadze_1D

I would suggest the Organic Chemistry Tutor on youtube, he is great!

Mashkinadze_1D

Hello, you would begin by balancing. Then you would use the two molar masses to convert both of the reagents to moles and continue to convert that based on the stoichiometric coefficients and molar mass of the product to see which creates less product. This is not too in detail, but I hope it helps ...

Mashkinadze_1D

To solve this problem we would do as follows: 1.20g CaCO3 x 1 mol CaCO3 /100.086g = 0.001199 mole CaCO3 or 1.20 x 10^-3 moles of CaCO3 0.155g CaSO4 x 1 mol CaSO4 / 136.134 g = 0.001138 mole CaSO4 or 1.14 x 10^-3 moles of CaSO4 ( (1.20 x 10^-3) + (1.124 X 10^-3) )/ (0.400) = 0.00585 M or 5.85 X 10^-3...

Mashkinadze_1D

To calculate this, you would have to look at the net charges of each molecule. This is a comprehensive topic but let me try to shorten it:\ All molecules must be neutral in charge Different atoms have different charges Group 1 elements have a charge of +1 Group 2 elements have a charge of +2 Group 1...

Mashkinadze_1D

To the extent of my knowledge, black bodies do not reflect light, only absorb. This being said, Professor Lavelle mentioned that this is much more so a physics concept and we should not worry heavily on memorizing or understanding it.

Mashkinadze_1D

Hello! In the majority of combustion, oxidation, and metabolizing reactions , carbon dioxide and water will be the products. Although these are usually the products, there may be additional molecular compounds that were also created during the combustion reaction. Hope this helps!

Mashkinadze_1D

Hello. Yes, the moles are representative of how many molecules are in a solution and therefore both of moles of a substance and the amount of molecules of such substance would remain the same during dilution. The only amount that would change during the dilution process would be the amount of moles ...

Mashkinadze_1D

Hello! The reason you are getting the problem wrong is a small error you made with the O2. When you divided for moles you found the moles of O2, but instead it would have been more helpful to find the moles of O which is 2.4475. You would then proceed to find K:O which would be about 2.98 or close e...

Mashkinadze_1D

Hello to solve this problem you would proceed as follows: 454g tristearin x 1mol/891.48g = 0.509 mol tristearin 0.509 mol tristearin x 163 mol O2/2mol tristearin x 32gO2/1molO2 = 1327.472 g of O2 but to correct sig figs is 1.33X 10^3g O2. The important part of this problem is using dimensional analy...

Mashkinadze_1D

Yes this is correct! M1V1 still equals the amount of moles and the amount of moles of a certain substance is not the same although the morality may be between two different solutions. So therefore we would use the new volume in the equation.

Mashkinadze_1D

To solve this you would want to proceed as follows: So assume that we have 100g of Xylitol to make this easier and the continue to 39.43 g Carbon x 1mol C/12.10 g C = 3.259 mol C 52.58 g Oxygen x 1mol O /16.00 g O = 3.286 mol O 7.88 g Hydrogen x 1mol H/1.008 g H = 7.82 mole H After this continue to ...

Mashkinadze_1D

To answer your question, I had the same idea at first and was a little confused but the molarity does not change. Molarity is in moles per liter and is therefore constant across a whole solution once all of the moles of the solute and solvent have been combined. Therefore, when we remove a certain a...

Mashkinadze_1D

Basing off the other two replies I want to add on that like they stated, the correct amount of sig figs will mostly depend on what is in the rest of the problem (previous sig figs, addition/subtraction, or multiplication/division) To expand in a situation where we are for example trying to reduce 12...

Mashkinadze_1D

I agree with the other user that the amounts should be leaning towards a value usch as 0.5 or 0.33. In addition, I would like to add that each step of the way to use the correct sig figs because this personally has helped me to find values that I can see where to round to more so. In many situations...

Search found 87 matches