CHEMISTRY COMMUNITY

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If you want practice with it, you have to use this approach in 7.23 (b) (7th edition) to find the rate law.

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This equation can also be used to compare k1 and k2, like in question 7E.3 (7th edition). But there will always be only one missing variable, everything else will be given.

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A higher (or more positive) E value means that the species will have a greater oxidation power. A lower (or more negative) E value means the species will have a greater reducing power. Just look at the tables and find the E values for the reduction reaction and rank them that way.

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In the back of the book, it says the answer for 7A.17 (7th edition) part (c) has the units of L^4 x mmol^4 x sec^-1, but shouldn’t mmol be to the negative fourth?

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For part (c) I did the calculation and got 2.85 but the book says that the answer is 2.85x10^12. I can’t figure out how to get to that answer. Am I supposed to do some sort of unit conversion during the calculation?

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When you are writing a cell diagram, how do you know to use a comma or a line on one of the sides to separate the reactant/products?

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In class Lavelle said that the anode (what is being oxidized) is always on the left and the cathode (what is being reduced) is always on the right.

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A salt bridge is when you have two different solutions that are in different beakers while a porous disc is used in a concentration cell in which you have two solutions that are the same but are at different molarities. Also, a porous disc is used when there is only one beaker while a salt bridge is...

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Can someone help with parts (c) and (d)? I'm kind of unclear about adding hydroxide and water and I can't really figure those two out.

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Voltage can be negative. In lecture, Lavelle was just saying that a battery cannot have a negative voltage or it will not work.

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The effect of temperature on a reaction depends on whether the reaction is endothermic or exothermic. If the reaction is endothermic, adding heat to it will favor the forward reaction and taking heat out of it will favor the reverse reaction. If the reaction is exothermic, adding heat to it will fav...

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If we are just given which ions are used in the reaction, how would we be able to tell which one is oxidized and which is reduced? For example, if a question just told us we had Cu^2+ and Zn^2+, how would we know that Cu is reduced and the Zn^2+ has been oxidized?

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Yes, that is why to find E°(cell) you subtract the anode from the cathode (therefore making the E° value for whatever is being oxidized the opposite sign).

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We write them as reductions because the EMF (E°) values of the half reactions are given for the reduction of those cations. If you were to use the oxidation reaction, you would switch the sign of the E° value (ex: the E° for reduction is 0.76V so the E° for oxidation is -0.76V).

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"Calculate the change in entropy when the pressure of 1.50 mol Ne(g) is decreased isothermally from 15.0 atm to 0.500 atm. Assume ideal behavior." I got the correct numerical answer for this, but why is the answer not negative? In the back of the book, it says that 42.4 J/K is the answer b...

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What exactly does it mean if someone were to say that H2O(l) "occupies more positions" than H20(s)?

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Heat Capacity at CONSTANT VOLUME Q = nCVΔT For an ideal gas, applying the First Law of Thermodynamics tells us that heat is also equal to: Q = (3/2)nRΔT + W, although W = 0 at constant volume. So, Q = (3/2)nRΔT. Comparing our two equations Q = nCVΔT and Q = (3/2)nRΔT we see that, for a monatomic ide...

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In an isothermal system, ΔS = [integral(T1 to T2) dq/T] Since q = nCΔT, dq = nCdT So, you can substitute the original equation to make it ΔS = [integral(T1 to T2) nCdT/T] You can pull the constants outside of the integral: ΔS = nC[integral(T1 to T2) 1/T dT] Using the general rule for integrals (the ...

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If a question gives you moles of a substance, it will probably give you the volume of the solution too. Use that to find the concentration then use the stoichiometric coefficients as the exponents in the equilibrium constant equation.

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I'm doing this problem but in the back of the book there is only one answer when the question asks for final temperature and change in enthalpy for two separate parts (parts a and b). Is this a mistake? What are the answers to this question?

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For part b, I got -1061.4 J. I used the equation -nRTln(Vf/Vi). I set n=1, 8.314 as the gas constant R, 305 K as my temperature T, my final volume as 6.52 L, and my initial volume as 4.29 L. However, the answer at the back of the book is -326 J. What did I do wrong? n does not equal 1. You have to ...

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This is what the textbook says (7th edition, page 484): "Except for sulfuric acid (and a few other rare cases), to calculate the pH of a polyprotic acid, use Ka1 and take only the first deprotonation into account; that is, treat the acid as a monoprotic weak acid. Subsequent deprotonations do t...

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Both K and Q have the same equation ([products]/[reactants]). K is the equilibrium constant. You use the equilibrium concentrations of the products and reactants to find it. Q is the reaction quotient. For Q, the reaction is not at equilibrium. You use Q to compare the current reaction to the reacti...

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If 10.04 is a pH/pKa value, then it has 2 sig figs. For pH/pKa/pKb, sig figs are counted as the amount of numbers past the decimal. So 10.04 would be 2 sig figs and 10.037 would be 3 (you count the zero because it is between two nonzero numbers).

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Temperature always changes K. If the reaction is endothermic (+ΔH), heating it will favor the formation of products (the forward reaction). If the reaction is exothermic (-ΔH), heating it will favor the formation of reactants (the reverse reaction).

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If the question gives you the initial/final concentration of some of the products/reactants but not ALL of them, you will need to set up an ICE table. You use it (and the quadratic equation) to find missing values for the concentration of each product/reactant.

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I worked out this problem and got that the equilibrium partial pressures of H2 and Cl2 are 8.39x10^-18 bar, but the book says that the answer is 3.9x10^-18 bar. I'm trying to find where I went wrong, does anyone know what I could have done?

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I've been working on this problem and can't figure out what I am doing wrong. I set up the ICE table and tried to solve for Kc, but I keep getting the wrong value. Can someone break this problem down for me?

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When you have an equation where all of the products and reactants are gases, is there any advantage to using Kp over Kc (or vice versa)? Which is better to use?

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When do we use bar and when do we use atm? Does it matter? What is the difference?

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O2- < N3- < Cl- < At- Polarizability is a measure of how easily an atom's electron cloud can be distorted. Larger/less electronegative atoms will more polarizable. Rb+ < Sr2+ < Na2+ < Be2+ Polarizing power is a measure of how easy it is to distort another atom's electron cloud. Smaller, more highly ...

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Yes, strong bases usually have weak conjugate acids and strong acids usually have weak conjugate bases. For example, HCl (strong acid) has the conjugate base of Cl- (weak base) and OH- (strong base) has the conjugate base of H2O (weak acid).

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The bond between fluorine and hydrogen is very strong because of very high electronegativity difference. Because of this, fluorine and hydrogen don't easily dissociate.

Strong acids are acids that completely dissociate. Since it is hard for HF to dissociate completely, it is considered a weak acid.

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In order to write the net ionic equation, the weak acid must be written as a molecule since it does not ionize to a great extent in water. The strong base that the weak acid is reacting with is fully dissociated. A strong base will essentially "force" the weak acid to become ionized. The h...

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You have to check the formal charges of each of the molecules. Oxygen is more stable when it has a double bond, so the SO4- molecule will prefer that structure.

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An induced dipole is when a (partial) positive or negative charge is brought near a nonpolar molecule and disturbs the arrangement of electrons in the nonpolar species. The electrons in this nonpolar species develop a dipole (one side has more electrons than another), so it is called an induced dipo...

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Dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the mol...

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Because energy is conserved, the energy of an incoming photon (if an electron is ejected from the metal surface) is equal to the energy needed to eject the electron plus the kinetic energy of the electron. In the equation E=hv, h is a constant, meaning the energy of the photon is directly dependent ...

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Ionization energy is the energy required to remove an electron from a gaseous atom or ion. Magnesium is in the second group and therefore is smaller than sodium DUE TO ADDITION OF A PROTON. This proton is positive and attracts the outer electrons CLOSER to the NUCLEUS which therefore means it is har...

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A single bond is a sigma bond. A double bond is a sigma bond plus a pi bond. A triple bond is one sigma and two pi bonds. A sigma bond is a basic covalent bond, with the bond in line with the bonding orbitals. You can only ever have one sigma bond between any two atoms. A pi bond is a covalent bond ...

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Dipole moments arise from differences in electronegativity. The larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment. The dipole moment is a measure of the polarity of the mol...

klarratt2

A single bond is a sigma bond. A double bond is a sigma bond plus a pi bond. A triple bond is one sigma and two pi bonds. A sigma bond is a basic covalent bond, with the bond in line with the bonding orbitals. You can only ever have one sigma bond between any two atoms. A pi bond is a covalent bond ...

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Repulsion strength is strongest between two lone pairs of electrons and weakest between two bonds (repulsion strength between a lone pair and a bond is in between these two). Therefore, lone pairs have a bigger effect on the bond angle than a bond.

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A dipole molecule is a molecule made of atoms that have significantly different electronegativities. In HCl, for example, Cl has a much higher electronegativity than H, so it will attract electrons more strongly. This creates a permanent dipole in which one side of the molecule (the Cl) is partially...

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HCl molecules have a dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has a slight negative charge. Since Cl is more electronegative, electrons are drawn to it more than they are to the H.

klarratt2

A single bond is a sigma bond. A double bond is a sigma bond plus a pi bond. A triple bond is one sigma and two pi bonds. A sigma bond is a basic covalent bond, with the bond in line with the bonding orbitals. You can only ever have one sigma bond between any two atoms. A pi bond is a covalent bond ...

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Atomic spectra are the transitions of electrons between electronic energy levels in isolated atoms. They are affected by interactions of the transitioning electrons with the nuclei spins and with the other electrons in the atom. Molecular spectra involve transitions in molecules with two or more ato...

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ml is the magnetic quantum number. This just tells you which orbital the electron is in. For example, an electron in chromium could either be in the -2, -1, 0, 1, or 2 orbital. ms is spin magnetic quantum number. This tells you whether the electron has up spin or down spin (the two electrons in an o...

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Will effective nuclear charge be on the midterm? I am a little unclear about what it is.

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I don't think it will! Our TA told us it will cover the photoelectric effect, the De Broglie wavelength equation, the Heisenberg uncertainty equation, atomic spectra, and quantum numbers. It will not cover electron configuration or multi-electron atoms.

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I'm not sure if it will be on this test, but on the formula sheet for last week it tells you how to convert between eV and joules.

1eV = 1.602x10^-19 J

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If we were following the regular rules, copper would be [Ar] 3d^9 4s^2. However, some copper is more stable when all of the d orbitals are full (meaning there are 10 electrons in that orbital instead of 9). In order for the d orbital to be full, though, the s orbital has to "lose" an elect...

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When do you use E=hv and when do you use E=pc? Are they interchangeable?

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For part (c), I used the equation c=(wavelength)(frequency) when frequency=2.5x10^16 (as given in the problem), and I got 12nm for the wavelength. However, in the back of the book, it says the solution is 8.8nm. What am I doing wrong?

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In this question, you are given energy in keV (140.511 keV). How do you convert between keV and joules?

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In the photoelectric effect experiment, UV light is shone on a piece of metal. Electrons are sometime ejected from the metal because of the UV light, which is picked up by a detector. Not all frequencies of light can cause electrons to be ejected. The equation E=hv tells us the energy of the incomin...

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I know that E=hv is for the energy of the incoming photon, but can Ek=1/2mv^2 also be used to describe the kinetic energy of the photon, or is it only used for the kinetic energy of electrons?

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The frequency of radiation is different for each metal because they each have different work functions. When each line intercepts the x axis corresponds to the threshold energy of that metal (where the energy to remove an electron from that metal equals the energy of the incoming photon, making the ...

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The H2O just means that it is copper (ii) sulfate hydrate, so you do add the molar mass of the 5H2O to the copper (ii) sulfate to solve.

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Having a 0.50 M NaOH solution means that there are 0.50 moles of NaOH in 1 liter of solution. So, you will need to dilute the 2.5 M NaOH solution until it is 0.50 M. You know that in the required solution of 0.50 M, there are 0.50 moles in 1 liter. You would divide this 0.50 moles by the molarity of...

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