CHEMISTRY COMMUNITY

RachaelKoh3A

Electron donating groups like CH3 basically destabilize the conjugate base anion because it donates its electrons such that the negative charge on COO- will intensify. This will ultimately make it less stable. Conversely, electron withdrawing groups will stabilize the conjugate base anion because it...

RachaelKoh3A

For acid-base reactions in which there is a weak acid or weak base, the textbook states to write the weak acid/base as the undissociated form in the ionic equation. This is because they dissociate to a small extent such that the dominant species in the solution will be its undissociated form. So, on...

RachaelKoh3A

Yes, thank you!

RachaelKoh3A

In the stabilization of oxoacid conjugates (like CCl3COO- or CBr3COO-) in which there are electron withdrawing atoms such as halogens, does it matter if the halogen is less electronegative than the oxygen atom?

RachaelKoh3A

Generally, the central atom is the atom that has the lowest ionisation energy. Ionisation energy increases across a period and decreases down a group.

RachaelKoh3A

i think AX2E has a bent shape with a bond angle that will be less than 120 degrees while AX2E2 also has a bent shape but with angles less than 109.5 degrees. This is on the basis that, AX2E comes from a trigonal planar shape with one corner of the triangle being a lone pair instead. Trigonal planar ...

RachaelKoh3A

Just to add on, lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond pair-bond pair repulsion. In the tetrahedral example mentioned by Kayla, the bond angle between the atoms in a trigonal pyramidal shape becomes less than 109.5 because the lone pair-bond pair repulsion is greater th...

RachaelKoh3A

I would assume so. Since the molecule will preferentially form the most stable structure.

RachaelKoh3A

I think knowing its exact bond dissociation energy is only helpful if you want to compare exact values or make calculations. But for predicting which is stronger, in general, ionic bonding will be stronger than dipole-dipole interactions.

RachaelKoh3A

SO2Cl2 has a tetrahedral shape. So it doesn't matter where you put the Cl or O atoms because it will be the same regardless.

RachaelKoh3A

Bond angles are formed when electron pairs repel each other. The electron pairs will assume a geometry that keeps them as far apart from each other as possible because of this repulsion.

RachaelKoh3A

I think the energy of the bonds have already been experimentally determined.

RachaelKoh3A

It depends on the kind of molecules/compounds you want to compare. For CHCl3 and CCl4 you are comparing the types of intermolecular forces of attraction between the molecules. CHCl3 is a polar molecule while CCl4 is a non-polar molecule. Dipole forces are the dominant intermolecular forces of attrac...

RachaelKoh3A

I think it depends on the electronegativity of the atom. For example in SO3-, the O will more likely take on the negative charge because it is more electronegative than S and hence will draw electrons more strongly to itself compared to S.

RachaelKoh3A

To add on, the elements in period 3 and below can exceed an octet because it has d orbitals. These d orbitals are energetically accessible and can hence have an expanded octet. Elements that are in the second period cannot exceed 8 electrons because there are only 2s and 2p sub-shells that can hold ...

RachaelKoh3A

1. Atomic radius generally increases down a group and decreases across a period. Down a group, valence electrons are in shells that are further away from the nucleus and hence atomic radius increases down the group. Across a period there is a decrease because effective nuclear charge increases and v...

RachaelKoh3A

The electron is added to an atom in its gaseous state!

RachaelKoh3A

A gas-phase atom is an atom in the gaseous phase.

RachaelKoh3A

The half filled 3d orbital is more stable. So 3d^5 4s^1 is more stable than 3d^4 4s^2 and hence results in the different electron configuration.

RachaelKoh3A

To add on, this happens because when the d orbitals are half or fully filled the atom/ion will be more stable!

RachaelKoh3A

Quantum numbers are basically used to describe an electrons position in the atom. So it is not the atoms that have quantum numbers but the electrons in the atom. Electrons in the atoms could have the same 4 quantum numbers for example the electron in the 4s orbital in either copper or potassium with...

RachaelKoh3A

The orbitals in Xe are filled in this order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d . So the 6s orbital has to be filled first, then the 4f orbital and then the 5d orbital. In Cr the orbitals are filled in this order: 1s, 2s, 2p, 3s, 3p, 4s, 3d. Since a half-filled 3d orbital is more...

RachaelKoh3A

E = hv is the energy of a photon. The energy of the incoming photon must match the work function in order for an electron to be removed from the surface of the metal. I think then E = hv would be the same as the work function.

RachaelKoh3A

1 m = 1 x 10^9 nm and conversely 1 nm = 1 x 10^-9 m.

So, 4.2 x 10^-7 m converted to nm will be 4.2 x 10^-7 x 10^9 nm which is just 420 nm after the powers cancel out.

Hope this helps!

RachaelKoh3A

I think it is because it isn't possible for a speed to be greater than the speed of light. So the electron's velocity being greater than the speed of light is impossible and would either mean it was done wrongly or the question might for example state that the model being tested is incorrect, like t...

RachaelKoh3A

With regards to the photoelectric experiment, the threshold energy is the minimum energy needed to remove an electron from its metal. In the equation: (E=hv) - threshold energy = (Ek = 1/2me-ve^2) E=hv is the energy per photon and this energy must be equal to or more than the threshold energy for an...

RachaelKoh3A

Hi! The mass of the crucible remains constant. We can find the mass of the product first. Mass of product = 28.35 - 26.45 = 1.90 g (2dp) Since a type of tin oxide was formed and the mass of tin in this tin oxide should not change, we can find the mass of oxygen in the product by subtract the mass of...

RachaelKoh3A

En = -(hR)/n^2 for n = 1,2, ... ∞) In the equation, R and h are both constants and so hR will remain the same for any value of n. When n becomes larger and larger, En becomes smaller and smaller. This is such that as n approaches ∞, En will approach 0. So if the energy level is very high, the energy...

RachaelKoh3A

I think Dr. Lavelle mentioned that the experiment needs to take place in a vacuum so that the electrons that are ejected will not react with particles in the air.

RachaelKoh3A

Sulfide refers to the metal sulfide of M which would have the formula MS. OH has a charge of -1 and so in M(OH)2 M has a charge of +2. S has a charge of -2 and so you need to find the Mr of MS.

RachaelKoh3A

Let Mr of M be x. % composition = 88.8% Form an equation in x using the formula, % composition = (mass of M / mass of M2O) x 100%. [2x/(2x+16.00)] x 100% = 88.8% Then use algebra to solve for x. 2x/(2x+16.00) = 0.888 2x = 1.776x + 14.208 0.224x = 14.208 Eventually you will get x = 63.4 g/mol Mr of M...

RachaelKoh3A

Hi! Since 1.000kg of AgNO3 is used, C6H9Cl3 is the limiting reactant as AgNO3 is in excess. Find the number of moles of C6H9Cl3 by taking 0.750 divided by the Mr 187.50 g/mol. Number of moles of C6H9Cl3 = 0.750/187.50 = 0.00400 mol The number of moles of AgCl produced is the same as the number of mo...

RachaelKoh3A

Hi! For Q9 the net ionic equation will cancel out all the ions that remained in the aqueous form. The molecular equation will have all molecules written in it. Eg. Cu(NO3)2 + 2NaOH -> Cu(OH)2 + 2NaNO3 However, to fin the net ionic equation it will look something like this. Rough equation: Cu2+(aq) +...

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Re: Why is HCOOH a stronger acid than CH3COOH? [ENDORSED]

Re: Ionic equation

Re: Conjugate of oxoacid

Conjugate of oxoacid

Re: Central atom

Re: Bond angle

Re: Less than Angles

Re: Does formal charge apply to the VSEPR Model?

Re: strength of bonds

Re: Atom Placement

Re: Bond Angels

Re: Energy of Bonds

Re: Boiling Point

Re: Formal Charge

Re: Octet Exceptions

Re: Periodic Trends in Relation

Re: Electron Affinity

Re: Electron Affinity

Re: exceptions

Re: Ion formation and unpaired e-

Re: Quantum numbers for the "exception" cases

Re: Electronic Config of tungsten (1E.13)

Re: Question 1B.15 Part B

Re: Problem 1.7 conversion of meters to nanometers

Re: Speed of Light

Re: Threshold energy

Re: Homework question L.35 part 1 [ENDORSED]

Re: Energy levels and electrons

Re: Vacuum

Re: 6th edition: E15

Re: HW Problem F.7

Re: Limiting Reactant Calculations

Re: Homework Question M9