CHEMISTRY COMMUNITY

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Electrons are an intensive property and therefore do not become affected by the other changing concentrations in the redox reaction.

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The test includes everything we've learned up until Friday's (2/22) lecture, which does include Gibb's Free Energy.

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The E value given is dependent on the reaction itself. Therefore, the concentration of the compounds involved in the reaction does not affect this value.

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Yes it will always be like that because when something is reduced, it gains electrons and when it is oxidized, it gains electrons.

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Since entropy is a state function and is dependent on other situations occurring in the reaction, we have to calculate each change in entropy separately and then add all of them together to find total entropy.

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The middle step accounts for the change in volume while the other steps only account for change in temperature which is why you have to add all the changes in entropy together in order to find the total deltaS.

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When calculating C, you have to use different R values, depending on whether the gas is monatomic or diatomic. The R value you should use is given on the equations sheet.

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When delta G is negative, the system(reaction) is spontaneous.

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When a reaction is isothermal, delta U (internal energy) equals zero.

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Reversible expansion does the most amount of work because the gas is pushing against the maximum possible external pressure and less of the energy is lost to heat.

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This equation is given for the ideal gas law so you'd only us it when you have a problem where the volume and moles of a gas are changing.

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Yes, bond enthalpy can be either positive or negative since the value is based on whether the formation of the bond requires or produces heat.

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In an isolated system, you assume that there are no interactions with the outside. Therefore, the heat in the system should not be considered.

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If the value of K is less than 10^-3, then we can assume that the value is too small to have any impact and therefore we can assume that X minus any number becomes just x.

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Phase changes are endothermic because, in order to go from one state of matter to another, the bonds have to be broken, which requires heat.

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A good way to memorize strong acids and bases is understanding the relationships between the two elements in terms of electronegativity. Usually, when they have very opposite electronegativities, the two elements form a strong acid/base.

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To be safe, it would be best to do all the calculations without rounding first, and then rounding to the correct number of significant figures once you have the final answer.

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Usually for equilibrium reactions that involve only gases, we use Kp to determine partial pressures.

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For the equilibrium constant K, it is not affected by any changes in pressure, temperature, or concentration.

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If the equilibrium constant K is larger than 10^3, then the equilibrium favors the product. And, if the equilibrium constant K is smaller than 10^-3, then the equilibrium favors the reactants.

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Are solids and liquids included in the equilibrium constant equation?

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Aqueous and gaseous products and reactants are included in the equilibrium constant but solids and liquids are excluded.

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Heterogeneous equilibrium is when there are different states of matter present in the reaction and homogeneous equilibrium is when all the states are the same.

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What is the difference between Bronsted acids and bases and Lewis acids and bases?

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HF has a stronger bond due to the higher difference in electronegativity therefore the bond is harder to break, meaning the proton is harder to donate. Therefore, HBr is a stronger acid since it has a weaker bond and can more easily donate a proton.

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Determine the VSEPR model for each individual atom and then determine what the bond angle would be based off of the model.

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Pi bonds are side side while sigma bonds are end to end. Therefore sigma bonds are inherently stronger. Also, when pi bonds rotate, they break, so they are weak.

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Count the number of electron densities that are surrounding the central atom. For example, if there are 4, then the hybridization is sp3. If there are 5, then it is sp3d, since the p orbital can only contain three, and s can only have one, the 5th electron goes to the d orbital.

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The rotation does not affect the strength of the pi bond. The pi bond simply is not capable of rotating, since it is a side-by-side bond, unlike sigma bonds, which are capable of rotating, since they are end-to-end bonds. Therefore, if the two carbon atoms in the compound were to rotate, this would ...

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Triple bonds are the shortest and single bonds are the longest because as more bonds between atoms occur, it brings the two atoms closer together.

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First, you should draw the lewis structures for both molecules. Then, you can see the types of bonds between each atom. Single bonds have one sigma bond, double bonds have one sigma bond and one pi bond, and triple bonds have one sigma bond and two pi bonds.

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How do you determine the difference between sigma and pi bonds?

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In order to compare melting points, you have to look at the different intermolecular forces that are found within each compound. For example, hydrogen bonds are much higher to break (melt) than dipole-dipole bonds. Therefore, if one compound has a hydrogen bond and another does not, the compound wit...

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Hybridization is the concept of mixing atomic orbitals into new hybrid ones.

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Hydrogen bonds occur between hydrogen and elements such as oxygen, nitrogen, and fluorine. Hydrogen bonds require a lot more energy to break apart. Since H2S does not have any hydrogen bonds, and H2O does, the melting point of H2O is much higher since it requires more heat to break the bonds apart.

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Hydrogen bonds are the strongest among the intra molecular bonds but are weaker than the intermolecular bonds (covalent and ionic)

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The VSEPR model for Lewis structures is based on electron pairs so you have to count them to determine which one they correspond to.

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Intensity does not affect the energy of the ejected electron. The intensity will only affect the amount of electrons ejected

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Degenerate orbitals mean that the electrons within the orbital contain the same energy level.

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To be safe, you should know what they are. Radicals occur when there is a single unpaired electron, making it highly reactive.

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There are no electrons in nodal planes because they are theoretical areas in between the lobes of orbitals that have zero probability of having electron density.

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For multielectron atoms after the atomic number 20, the 4s state is higher energy than the 3d state, which is why it is written first in electron configurations. Also, the first electron to be removed is the lower energy levels so, 4s.

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Since one electron has already been removed, the remaining electrons are now experiencing a greater attraction to the nucleus. The subsequent electrons being removed will have a higher ionization energy since they are less easy to remove.

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Excited states of electrons have higher energy levels and the higher level energy electrons naturally want to move to the ground level of electrons.

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The equation sheet has all the equations that we need but they aren't labeled so you need to be able to know which equation is which and when to use each one.

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Most of the equations that we need to know for Test 2 are on the equations sheet in the front page of the test. But you should probably know what they are used for, what the variables represent, and how to manipulate them.

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There are some elements that have g and h orbitals, but only in their respective excited states. There are no discovered ground elements with g or h orbitals... and I don't think we'll need those concepts for this class.

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The energy "E" from the equation is the initial energy of the photon that is responsible for removing the electron. The energy "released by the photon" is most likely referring to the small amount of energy that the electron emits when it is removed and moves with some velocity.

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The DeBroglie equation is used in determining if the object being observed is showing wave-like properties. If the wavelength of the observed object is higher than a certain threshold, then the object is showing wavelike properties. So, you would use this equation when finding the wavelength of a mo...

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You should answer each part with the amount of sig figs that correspond to that part. For example, if in Part A, the problem gives values with 2 sig figs, then answer Part A in 2 sig figs.

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The actual yield refers to the amount of product that is produced from a chemical reaction after taking the limiting reactant in an equation into account. This differs from the theoretical yield, which would be the amount of product produced assuming no limitations.

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For the final answer, you should use the same amount of significant figures that are given in the problem. Also, make sure not to use significant figures while showing your work; you should only round to significant figures when you are producing your final answer.

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