CHEMISTRY COMMUNITY

Stuti Pradhan 2J

A good rule to follow is when the K value is less than 10^-4. This means that there is a lot of the reactant present, so any small change will not make a significant impact on the concentration of the reactant. However, in this question the K value is much larger than that, so you cannot use the app...

Stuti Pradhan 2J

If the salt has the conjugate acid of a weak base, the solution will be acidic because the conjugate acid will attract the OH- from water and leave H+ ions behind, making the solution acidic. It is also important to note that the other ion (in this case the anion) must be a spectator ion in this sit...

Stuti Pradhan 2J

For #5 on the week 2 Sapling hw, the question is as follows: The Kb for an amine is 5.266×10−5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.572 ? Assume that all OH− came from the reaction of B with H2O. I was wondering how to tell that this solution is at eq...

Stuti Pradhan 2J

First, convert the Ka value to Kb using the equation Kw=Ka x Kb where Kw is 1.0 x 10^-14 since the salt acts as a weak base. Then set up an ICE table for the equation ClO- + H2O <--> HClO + OH-, using 0.056 as the initial value. Set the Kb value equal to [HClO][OH-]/[ClO-], and solve for the OH- con...

Stuti Pradhan 2J

For this problem, I recommend converting the pH to a pOH since you are working with a base. Once you have a pOH, you can use it to figure out the OH- concentration. Since for every one mole of OH- there is also one mole of BH+, you now know the concentration of OH- and BH+. Set up the equation for K...

Stuti Pradhan 2J

For this question, it is important to know that the cation of a strong base and the anion of a strong acid create a neutral salt, the cation of a strong base and the anion of a weak acid create a basic salt, and the cation of a weak base and the anion of a strong acid create an acidic salt. HClO4, H...

Stuti Pradhan 2J

For this problem, I recommend converting the pH to a pOH since you are working with a base. Once you have a pOH, you can use it to figure out the OH- concentration. Since for every one mole of OH- there is also one mole of BH+, you know can figure out the numerator of the equation for Kb. Set the Kb...

Stuti Pradhan 2J

You typically use an ICE table when you are given initial concentrations. The question should state if the concentration is at equilibrium, or say that the initial values are given. If the concentrations of all the species in the reaction are given and you are asked to find the K value, those values...

Stuti Pradhan 2J

You basically just combine equations to get the equation you want. Equations can be reversed, in which case the K value would now be equal to 1/K, and the equation can be multiplied by a number, in which case the K value would be raised to the power of that number. For this question, multiplying the...

Stuti Pradhan 2J

If the K value is less than 10^-4, that means the reactants are strongly favored and x will not significantly change the concentration of the reactants, so the approximation will work. I believe for this class most cubic equations will have small K values where the approximation can be applied, but ...

Stuti Pradhan 2J

I think the main concept to understand with endothermic and exothermic reactions is that endothermic reactions require heat to form the products and exothermic reactions release heat. If heat was added to an endothermic reaction, then more products would be formed (product formation would be favored...

Stuti Pradhan 2J

The reaction quotient is used when the reaction is not at equilibrium. At equilibrium, the ratio or products to reactants is the K value, but when the reaction is not at equilibrium, there can be more products or reactants, so the ratio would no longer equal the K value. When a product or a reactant...

Stuti Pradhan 2J

I think Dr. Lavelle meant that since there is now more heat being added, that additional heat will be used by the reaction, which will result in more products being formed since the reaction is endothermic and needs heat to convert the reactants to products.

Hope this helps!

Stuti Pradhan 2J

For this problem, you can take the square root of both sides after setting up the ratio, so you do not need to solve a quadratic equation. Typically, if you have two positive concentrations, the one that is larger than the initial concentrations is incorrect.

Hope this helps!

Stuti Pradhan 2J

You are correct in that Q would be less than K, but K would not change. The equilibrium constant, K, only changes for a reaction due to a change in temperature. Therefore, not having reached equilibrium does not change the equilibrium constant. Calculating Q in this situation would let us know that ...

Stuti Pradhan 2J

Partial pressure is the pressure of one of the gases in a mixture. Essentially, the pressure of one gas if it was placed alone in the same volume as the mixture, is its partial pressure in the mixture. Since there is a mixture of gases, each gas has its own partial pressure, which can be added to fi...

Stuti Pradhan 2J

Solvents aren't written in the equilibrium constant equations because there tends to be such an excess of solvent, that even if there was a small change, it is insignificant. For example, if there are 1 million bees and 10 more are added, we still say there are about 1 million bees because the chang...

Stuti Pradhan 2J

K only changes due to a change in temperature. In this situation, adding more reactants or products would change Q because the concentrations are no longer at equilibrium. You could then make an ICE table and find the new equilibrium concentrations, which would still have the same ratio and therefor...

Stuti Pradhan 2J

The ideal gas law, PV=nRT is only used to convert from pressure to concentration, or vice versa. You always use K=[P]/[R] to find the equilibrium constant, but if you were given some values of the products or reactants as concentrations and other values as pressures, you can use the ideal gas law to...

Stuti Pradhan 2J

An ideal gas is based on certain assumptions of the Kinetic Molecular Theory. For an ideal gas, it is assumed that the molecules are constantly in motion, they have negligible volume, they do not exert any force on each other, and that they have perfectly elastic collisions. These assumptions allow ...

Stuti Pradhan 2J

I believe you are correct. The way I think of it is that anything with lower energy is more stable, so in an endothermic reaction the products would have more energy than the reactants and therefore, be less stable. On the other hand, in an exothermic reaction, the products would have a lower energy...

Stuti Pradhan 2J

I believe the conjugate acid would just be water in this situation, or 2 water molecules since the equation has (OH)2. Your second equation with the conjugate base of the weak acid is correct as that would be the reason as to why the pH increases to above 7.

Hope this helps!

Stuti Pradhan 2J

How do you use valence bonding descriptions when it comes to triple bonds? For example, you use valence bonding descriptions such as (C2sp^2, C2sp^2) to describe a bond between two carbon atoms with trigonal planar geometries, but in HCN, how would you describe the triple bond between the C and the ...

Stuti Pradhan 2J

The easiest way to do this problem is to simply choose a molar concentration of HCl and find the pH. Then, take 12% of that molar concentration and find the pH value. Subtract the values to find the difference in pH. For example, I used a 0.5 M solution of HCl, and then a 0.06 M solution (which is 1...

Stuti Pradhan 2J

In Cu(NO3)2, the NO3- is the conjugate base of a strong acid (HNO3), so it will not affect the pH of the aqueous solution. However, Cu2+ is not the conjugate acid of a strong base, so it will affect the pH. Since Cu2+ has a positive charge, it will attract the oxygen atom from the water molecule. Th...

Stuti Pradhan 2J

In this problem, you want to see what the effect of the salt would be on the aqueous solution, so you want to understand how the salt will interact with the water molecules. Within the salt, the conjugate base of any strong acid and the conjugate acid of any strong base will have no effect on the pH...

Stuti Pradhan 2J

T-shaped does not have lone pairs in the axial position (top and bottom) but instead, has them on the equatorial position. This is because when there is only 1 lone pair, making the atom on the top a lone pair would cause the more repulsive lone pair to interact with three atoms (all the atoms in th...

Stuti Pradhan 2J

You want to see what the effect of the salt would be on the aqueous solution, so you want to understand how the salt will interact with the water molecules. Within the salt, the conjugate base of any strong acid and the conjugate acid of any strong base will have no effect on the pH, so that ion can...

Stuti Pradhan 2J

Why does the acid deprotonate when the pka<pH?

Stuti Pradhan 2J

You want to begin by naming the ligands in alphabetical order. Since you have ammonia and chloride, ammonia will come first. You need to specify the number of that ligand, so it would be triammine. You also need to specifiy the number of the chloride, so it would be trichloro. Finally you add the na...

Stuti Pradhan 2J

The coordination number is the number of bonds formed to the transition metal, so in this case, there are 5 NH3 molecules and 1 SO4 molecule that are bonded to the cobalt, resulting in a coordination number of 6. I recommend first counting how many ligands there are (so anything that is within the c...

Stuti Pradhan 2J

The potassium should go at the beginning (before the brackets) because it is a cation. Anions, such as Cl- or F- go after the rest of the name. This is the same idea as you would see in any salt, where the convention is to write the cation first and then the anion (ex. NaCl). In the case of potassiu...

Stuti Pradhan 2J

The weaker the bond in an acid, the stronger the acid is because it can dissociate easily. So, if the molecules are particularly large (have a large atomic radius), then it is more likely that it is a strong acid. There are not a lot of strong acids, so it seems likely that anything not included on ...

Stuti Pradhan 2J

Using the same set up as the H+ will give you the pOH. The equation is pOH=-log[OH]. To find pH from this, you can use pH+pOH=14. Otherwise, you can use the equation [H+]=1.0 x 10^-14/[OH-] to find the H+ concentration, and then find the pH from there. This diagram is helpful to see how to convert b...

Stuti Pradhan 2J

To figure out if compounds are in the same plane or not, you have to look at the shape of the compound. Compounds that are linear, trigonal planar, bent, t-shaped, and square planar are all coplanar, which can be seen by their molecular geometries which have all the atoms in the same plane. Hope thi...

Stuti Pradhan 2J

Cyanide does have 2 lone pairs, but since it has a triple bond and is linear, the molecule cannot orient itself in a way the both lone pairs bond to the same transition metal. To have a polydentate ligand, you need at least two lone pairs and at least one spacer between the two lone pairs. Cyanide d...

Stuti Pradhan 2J

n=2 only has s and p orbitals, since the first quantum number possible for a d orbital is 3. There is 1 s orbital and there are 3 p orbitals, so n=2 has 4 orbitals.

Hope this helps!

Stuti Pradhan 2J

C3H4 has three resonance structures. C3H4.png In the first structure given in this image, the carbon on the left has two regions of electron density and is therefore, linear and sp hybridized. This has a 180 degree bond angle. The carbon atom in the center has a triple bond and a single bond, which ...

Stuti Pradhan 2J

To answer the first part of your question, yes, all molecules with resonance have unhybridized molecules. Since resonance typically implies that there is a double bond, there must be an unhybridized molecule to create the pi bond. In terms of hybridized orbitals, this site has a lot of diagrams that...

Stuti Pradhan 2J

Hybridization does correlate to specific electron geometries, since it is based on the number of regions of electron density around the atom. So, for example, all molecules with linear electron geometry would have an sp hybridization, all molecules with trigonal planar electron geometry would have a...

Stuti Pradhan 2J

From my understanding, there has to be an unhybridized orbital in order for a \pi bond to form. I was wondering if this had to be a unhybridized p orbital or if it could be an unhybridized d orbital as well. Additionally, in sulfuric acid, H2SO4, the central S atom has an sp3 hybridization. However,...

Stuti Pradhan 2J

In today's lecture, Dr. Lavelle said that in a sp2 hybridization (in the molecule C2H2 for example) the fourth electron of the carbon would have to be in the unhybridized 2p orbital because the energy difference between the sp2 orbitals and unhybridized p orbital is very small. I know it has somethi...

Stuti Pradhan 2J

In the first example of today's lecture regarding the hybridization of ammonia, Dr. Lavelle says that we need to have sp3 hybridization because a bonding model with just the normal s and p orbitals would mean that the three unpaired electrons in the p orbital would have 90 degree bond angles, which ...

Stuti Pradhan 2J

Are central atoms the only atoms in a molecule that can undergo hybridization? Why or why not?

Thank you in advance!

Stuti Pradhan 2J

Hybridization occurs to create hybrid orbitals that can better accommodate the pairing of electrons. The numbers of regions of electron density can tell you the hybridization. For example, a molecule with only 2 regions of electron density would have an sp hybridization because it would only involve...

Stuti Pradhan 2J

It is linear because if there was one lone pair, and four bonding pairs, the lone pair would be in the equatorial plane so the more repulsive lone pair only reacts with 2 bonds at 90 degrees. Using the same logic, the next two lone pairs would also be removed from the equatorial plane, leaving two b...

Stuti Pradhan 2J

Yes, l=5 is a valid answer. There are other orbitals past the s, p, d, and f orbitals (for example there is a g orbital that exists, but there are no elements that have been found that can occupy this orbital). Any orbital that satisfies the 0, ..., n-1 requirement is a possible value for l. Hope th...

Stuti Pradhan 2J

Since the s orbital is around the nucleus, s electrons are able to shield outer electrons well. On the other hand, p orbitals have a node where the nucleus is, so they are not able to shield outer electrons well. I also believe that d orbital electrons are not able to shield electrons very well eith...

Stuti Pradhan 2J

From my understanding, since iodine has a larger atomic radius and therefore, its valence electrons are less tightly held, it is more polarizable. This will result in stronger LD forces between CHI3 molecules than CHF3 molecules, which require more energy to break and thus have a higher normal melti...

Stuti Pradhan 2J

Yes, I believe the entire column under Cr and Cu also follow that same rule. However, I do not think Dr. Lavelle will ask questions related to electron configurations for those d block elements.

Stuti Pradhan 2J

Those elements themselves in the ground states do have electrons that occupy the 4p orbitals. However, since its As3+ and Ge2+, that means that 3 of the electrons have been removed from As and 2 of the electrons have been removed from Ge. The 3 electrons removed from As to get As3+ all come from the...

Stuti Pradhan 2J

The ground state for Ge is Ge:[Ar]3d10 4s2 4p2, so for Ge2+, it would be Ge2+:[Ar]3d10 4s2, since two electrons are removed. The electrons must be removed from the highest energy shell, which is why they are removed from the 4p orbitals.

Hope this helps!

Stuti Pradhan 2J

The difference in electronegativity between the hydrogen and the carbon is not enough to give the hydrogen a significant partial positive charge, so it is not attracted to the lone pair electrons on other oxygen, nitrogen, or fluorine atoms. Hydrogens attached to carbons mainly only have induced dip...

Stuti Pradhan 2J

The hydrogen bonds push the molecules further apart, which leads to ice having a lower density than water, causing it to float. It is not so much the shape of the molecules, but more their spacing due to the hydrogen bonding.

Hope this helps!

Stuti Pradhan 2J

Resonance structures mainly occur when there is a double bond that can be at multiple locations in the Lewis structure, while still following all the other guidelines, such as making sure every atom has an octet. I do not think there are any specific steps you need to follow to determine resonance s...

Stuti Pradhan 2J

To find the most favorable resonance structure you should look at the formal charge on all the atoms. A compound where each atom has a formal charge of 0 is the most stable and therefore will contribute the most to the resonance structure. Other things to look out for is that you do not want highly ...

Stuti Pradhan 2J

There are two main requirements for forming hydrogen bonds. The first is that either an O, N, or F molecule must have a lone pair that can become the hydrogen bond. The second is that the hydrogen must be partially positive (), in order for it to be attracted to the lone pair. The partial positive c...

Stuti Pradhan 2J

In this diagram, the double bond between the nitrogen and the oxygen can technically be between the nitrogen and either oxygen, which causes resonance. When there is resonance, it is not that there is one set of double bonds and another single bond, but instead, both nitrogen-oxygen bonds are in bet...

Stuti Pradhan 2J

I typically look at what it is bonding with to figure it out. For example, in the question you are referring to, the AlCl3 reacts with Cl-, which donates a pair of electrons. Al is also a metal and a cation with an empty valence shell, so it will most likely accept electrons. On the other hand, Cl i...

Stuti Pradhan 2J

An easy way to tell the difference is that ions tend to lose of gain electrons resulting in charges, such as Na+, or Cl-, whereas a dipole indicates an unequal sharing of electrons due to differences in electronegativity, resulting in some partial charges. If the difference in electronegativity is g...

Stuti Pradhan 2J

The valence electrons just depend on the periodic table, so nitrogen would have 5. In the Lewis structure, you just want to make sure that nitrogen has a full octet. This means that it has 8 electrons, either as bonds or lone pairs. Each lone pair has two electrons and each bond has two electrons. T...

Stuti Pradhan 2J

There are two main requirements for forming hydrogen bonds. The first is that either an O, N, or F molecule must have a lone pair that can become the hydrogen bond. The second is that the hydrogen must be partially positive ( \delta + ), in order for it to be attracted to the lone pair. The partial ...

Stuti Pradhan 2J

The formula for formal charge is the number of valence electrons minus the number of bonding pairs (or the shared electrons/2) minus the number of lone electrons. In the scenario where there was a single bond between the oxygen and the sulfur, the formal charge would be FC= valence electrons - bondi...

Stuti Pradhan 2J

A resonance structure shows a possible bonding arrangement, and there can be multiple resonance structures that all show possible arrangements of the bonds in the molecule. However, a resonance hybrid is an average of all the resonance structures, and it more accurately depicts the relative structur...

Stuti Pradhan 2J

The closer the given bond length is to the measured length of the double or single bonds, the more it exhibits the character of that specific bond. For example, since the carbon-nitrogen bond is 136 pm, which is closer to the length of the carbon-nitrogen double bond (127 pm), it has more of the dou...

Stuti Pradhan 2J

Formal charge just gives you the most stable structure for any molecule. In the case of BF3, if there was a double bond, fluorine would have a formal charge of +1. Fluorine is very electronegative and has a high electron affinity, so a +1 charge, which would indicate only 6 valence electrons instead...

Stuti Pradhan 2J

n=5 does exist already, but the corresponding l values (such as l=4 which would be an g orbital) would be occupied by an element that has yet to be discovered. I think for any element to have an electron that occupies the l=4, you would need element 121, which has not been found yet.

Stuti Pradhan 2J

Any element that exists in 3p orbitals or greater can have an expanded octet. This is because there is an unoccupied d orbital which can be used for bonding. Si, P, Cl, and S are all common examples of elements that can have expanded octets (as they all have electrons in the 3p state). Additionally,...

Stuti Pradhan 2J

I think there are g, h, and i orbitals, there are just no elements with electrons at a ground state in any of those orbitals. The first element that has an electron that would occupy the g orbital would be element 121, which has not been discovered yet.

Hope this helps!

Stuti Pradhan 2J

Delocalized electrons are basically electrons that are not associated with a specific atom, but instead are equally likely to be located at any of the bonds involved with the resonance structure. For example, in benzene (C6H6), the electrons are delocalized, meaning that there is an equal probabilit...

Stuti Pradhan 2J

I am not sure exactly what question you are referring to, but if a colored line is produced, that means it is part of the Balmer series, so n=2. If the question does not specify anything about the type of radiation emitted, you can check if the wavelength is part of the visible light region (420 to ...

Stuti Pradhan 2J

The equations E=hv and c=vλ refer to electromagnetic radiation and therefore can only be used for photons. On the other hand, λ=h/p and ke=1/2mv^2 are for any particles that have rest mass. The second two equations can be used for electrons, neutrons, protons, etc. Since photons have no rest mass, t...

Stuti Pradhan 2J

The reason the oxygen is unknown is because combustion reactions typically have a compound that reacts with an excess amount of oxygen to form carbon dioxide and water. Since all the carbon in the carbon dioxide and all the hydrogen in the water are from the original compound, you know that the mass...

Stuti Pradhan 2J

To find n2, convert the wavelength into meters, and then use the equation c=\lambda \nu to find the frequency. Once you find the frequency, you can use the Rydberg equation. Input the frequency, Rydberg's constant, and the final energy level into \nu = R (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&...

Stuti Pradhan 2J

To find n2, convert the wavelength into meters, and then use the equation c=\lambda \nu to find the frequency. Once you find the frequency, you can use the Rydberg equation. Input the frequency, Rydberg's constant, and the final energy level into \nu = R (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}&...

Stuti Pradhan 2J

In the lecture, Dr. Lavelle explains how one of the anomalies in the ionization energy, why oxygen has a lower ionization energy than nitrogen, is because oxygen has 4 electrons in the 2p orbital. This means that the 2px orbital has two electrons, unlike nitrogen, which only has one electron for eac...

Stuti Pradhan 2J

I had this same question, so I asked a TA about it. It is true that each intermediate change in energy would cause an individual spectral line, so there should be more intermediate lines. However, for the sake of the question, Sapling tried to keep it simple, or at least that is the conclusion we ca...

Stuti Pradhan 2J

Trailing zeros are not significant when there is no decimal point. For example, 200 would only have 1 sig fig because the trailing zeros are insignificant. However, since 2.000 x 10^1 has a decimal point, the zeros are significant. This is because 200 could be a rounded number or there is some other...

Stuti Pradhan 2J

I would assume that you could use E=h\nu as you would with the frequency of a single photon because the frequency per photon gives you the energy per photon, so the frequency per mole should give you the energy per mole. The Js^-1 unit for Planck's constant should cancel with the J for the energy, a...

Stuti Pradhan 2J

In this situation, they are giving you a wavelength and asking you to find the kinetic energy. These are related by the DeBroglie equation, so you first need to use that to find the velocity of the electron. Once you figure out the velocity, you can input that value into the kinetic energy equation ...

Stuti Pradhan 2J

Your new equation looks correct. I used the equation you listed and got 6.7x10^-26 J, so maybe check your calculations because all your work before that looks correct. Make sure you converted the wavelength into meters, that could just be a small error.

Hope this helps!

Stuti Pradhan 2J

Yes, chlorine gas is Cl2. There are 7 diatomic molecules (H,O,N,Cl,Br,I,F), so whenever these are gases or "exist naturally", you should use the subscript 2.

Hope this helps!

Stuti Pradhan 2J

Most of the energy equations require kg because Joules = kg⋅m^2⋅s^−2, so you need to use kg to make sure the units cancel. In converting from mass to moles, you just use the molar mass in grams, but otherwise, kg it typically used as it is the SI unit for mass.

Hope this helps!

Stuti Pradhan 2J

The main thing to remember here is that 4 quantum numbers specify a single electron. Therefore, n=6,ℓ=2,mℓ=−1, which has three quantum numbers and simply does not include the magnetic spin quantum number is referring to two electrons that are in the sixth shell in the d orbital. The mℓ number tells ...

Stuti Pradhan 2J

The way to balance this equation is: 3NaHCO_{3}+C_{6}H_{8}O_{7}\rightarrow Na_{3}C_{6}H_{5}O_{7} + 3H_{2}O + 3 CO_{2} I usually approach these problems by balancing any elements that are not hydrogen or oxygen first. In this case I began with Na. After balancing the Na, I balanced the C, and then th...

Stuti Pradhan 2J

If given the wavelength of an ejected electron, you would use the DeBroglie equation because the electron has mass. You could figure out the velocity and then from there figure out the kinetic energy and relate that back to the incident light, depending on what the question asks for. You are correct...

Stuti Pradhan 2J

The energy required to remove an electron is the work function because you could have incoming light with greater energy than is required to remove the electron, which would then give you an ejected electron with kinetic energy (think about the equation E(photon)-work function= KE). The energy that ...

Stuti Pradhan 2J

Is it possible for electrons to fall to intermediate energy levels? For example, could an electron that is excited to the energy level n=5 fall to n=3 before returning to the ground state? Additionally, since each of the changes in energy levels would have a different change in energy (such as n=5 t...

Stuti Pradhan 2J

Since you are told that 1 meter is 1,650,763.73 wavelengths of radiation emitted by krypton-86, it is basically a 1 meter to 1,650,763.73 wavelengths ratio. To find 1 wavelength, just divide 1 meter by 1,650,763.73. If you set it up as 1m=1,650,763.73 wavelengths, then your goal is to find 1 wavelen...

Stuti Pradhan 2J

The Paschen, Brackett, and Pfund series are all in the infrared region. I believe that their lowest energy levels are n=3, n=4, and n=5, respectively. I am assuming we will not be required to know these though.

Hope this helps!

Stuti Pradhan 2J

Since the kinetic energy is the difference between the energy of the photon and the energy required to emit a photon, you should figure out the energy that corresponds to a wavelength of 194 nm to begin 34B. Subtract the energy you found in 33A (the work function for Molybdenum) from the new energy ...

Stuti Pradhan 2J

The values should be the same, it just depends on what information you are given/need to find. If you are dealing with momentum (mass and/or velocity) then you should use DeBroglie's equation. If you are dealing with energy and/or frequency, you should use one of the other two.

Hope this helps!

Stuti Pradhan 2J

I would convert the wavelength into meters and then find the frequency. Once you have the frequency, plug it into Rydberg's equation. Since it appears as a blue line, we know that it is part of the visible spectrum, so the Balmer series. That means that n1=2. From there you should be able to solve f...

Stuti Pradhan 2J

Yes, that is correct! Think about it like (5.11x10^-4 cm) x (10^-2m/cm). That way the units cancel and you get the right conversion.

Hope this helps!

Stuti Pradhan 2J

I got C49. I found the mass percentages and then found the moles of C out of 100. g to get 5.20 mol C. Then, I divided by the smallest amount of moles (0.637 moles N) to get 8.16 as the ratio of C. When I multiplied that by 6, I got C49. Maybe your ratio or one of the intermediate calculations was s...

Stuti Pradhan 2J

It is 6.626x10^-34 Js.

Stuti Pradhan 2J

I got the same answer for that question. I would assume that a difference that small simply comes from rounding (maybe for Planck's constant or pi).

Hope this helps!

Stuti Pradhan 2J

You have to use meters because the other values are are in meters and ultimately, the units need to cancel. For example, the speed of light is 3.00E8 m/s. To make sure the units cancel in the equation c=\lambda \nu , you would need your wavelength in meters as well, since the frequency is in Hz, or ...

Stuti Pradhan 2J

The mass tends to have a larger impact on the wavelength. If you want to test it out, you can find the wavelength of a baseball traveling at 1 m/s (or any very slow velocity). The wavelength is still way too small to be detected. Since Planck's constant is such a small number, the mass needs to be v...

Stuti Pradhan 2J

It is hard to say exactly how large an object can be since DeBroglie's equation depends on both the mass and velocity. I would say anything that you can hold is too large. If you ever need to check, you can always just input the information into the equation and any object with a wavelength smaller ...

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