CHEMISTRY COMMUNITY

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The line separates the electrode.

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No. Cell diagrams are simply a summary of the players in the redox reactions.

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If it is the solvent, then you can leave it out of the rate law. This is because solvents are present in such high concentrations that their concentrations don't change much during the reaction.

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If there are any H+ ions, you need to convert them to H2O using OH-.

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Enthalpy is a state property and entropy is not.

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Using the 8.314 results in the correct units.

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You can also use the formulas given in the equations sheet. Just make sure you understand the formulas and which order reaction to use them for.

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The rate is independent of the concentration if the reaction is zero order with respect to that reactant.

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The rate equals k(concentration of H2)(concentration of I2). It is, in a way, k(concentration)^2.

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Reducing power is the ability of something to give up electrons so that those electrons can go reduce some other molecule.

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The terms themselves imply the sign. If you have a negative rate of consumption, you would be forming the molecule; it doesn't make sense to call it the rate of consumption. The same goes for rate of formation.

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The lower the standard reduction potential, the more unlikely the metal will get reduced (since more negative E values result in more positive G values).

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For this question, the overall equation is given as Cd(s) + 2 Ni(OH)3(s) --> Cd(OH)2(s) + Ni(OH)2(s). However, in the cell diagram, there was a KOH. Where did the potassium come from?

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What are the rules for balancing a redox equation with only one reactant/product? For example, the equation given in 14.13.d is Au+(aq) --> Au(s) + Au3+(aq). I thought Au+(aq) would be used in both half reactions but the solutions manual chose to use Au(s) for both reactions instead.

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Why is the "n" in this equation always positive even when there is a negative change in moles?

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deltaG = deltaH - TdeltaS. If deltaG is negative, then the reaction is spontaneous. Since deltaH is negative, the reaction will be spontaneous for all positive values of deltaS and small negative values of deltaS.

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They are asking for molar entropy, so there are 6.02E23 molecules.

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The question asks for molar entropy, and there are 6.02E23 molecules. 6 orientations and 6.02E23 molecules results in 6^6.02E23. Plug it into the Boltzmann's equation and you will get the answer.

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It depends on the temperature.

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The problem gives all reactants and products. As for the phases, everything is in gas form at the given temperatures. Water can be liquid, but the problem specifies that the reaction uses water vapor.

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The method you mentioned should work as long as you make sure that you end up with the units that correspond to the variable you are solving for.

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Use the specific heat capacity of ice when you are raising the temperature of ice. Because the ice is already at 0C, you don't need to use the specific heat capacity of ice because the ice doesn't get warmer than 0C. Instead, use the heat of fusion to calculate the energy needed to melt the ice, the...

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If the equation requires deltaT, then it doesn't matter whether you use K or C. If the equation requires T and not deltaT, then usually you will need to use K. Make sure to keep track of your units and that after cancelling, your ending units are actually what you are looking for.

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The values given in the appendix are accurate to the hundredths place. When you add them, your answer will still be accurate to the hundredths place. It happens that with the hundredths place, you will have 6 sig figs.

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deltaU = q + w. When no work is done, U=q. When no heat is transferred, U=w.

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For irreversible reactions, w=-(Pext)(deltaV). For reversible reactions, w=-nRTln(v2/v1).

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In an isolated system, neither particles nor heat can be transferred. In a closed system, heat can be transferred but particles cannot.

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The solution stops at 2NO(g) + 3/2O2(g) --> N2O5(g) because you can use the enthalpy of formation of NO(g) to figure out the enthalpy of formation of N2O5(g). For 2NO(g) + 3/2O2(g) --> N2O5(g), deltaH is -169.2 kJ/mol. Thus, 1(deltaHfN2O5) - 2(deltaHfNO) = -169.2 kJ/mol. Using the appendix, 1(deltaH...

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A reversible isothermal reaction is one in which the volume changes slowly at a constant temperature. There is no heat transfer, so q=0, so deltaU=w.

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For homework problem 8.37b, we are asked to calculate the enthalpy of vaporization of ethanol (given that 21.2kJ vaporized 22.45 grams). The answer key gives the answer as 43.5kJ/mol, but would 0.944kJ/g also be correct? In other words, what units are we supposed to use for enthalpy of vaporization ...

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We are using the equation M1V1=M2V2 because the number of moles is the same between the two solutions. Thus, (0.18M)(0.5000L)=M2(0.00500L). The solution manual simply rearranges this.

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From the problem, we know that the concentration of HCl in the intended solution is 0.025M. Therefore, [H3O+] in the intended solution is also 0.025M. The number of moles of HCl remains the same between intended and actual solutions. From earlier in the course, we learned that M1V1=M2V2. Thus, using...

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The nitrogen is already very unhappy having 9 outer shell electrons in the single bond. I imagine that it will become suicidal if you decide to give it 10 outer shell electrons with the double bond. It wasn't even supposed to have an expanded octet in the first place. Don't force the relationship or...

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Simply changing the concentration of the reactants or products will not change the equilibrium constant. If you add more reactant, then the reaction will shift to the right. If you add more product, then the reaction will shift to the left. In the end (after you wait a while for the reaction to reac...

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When carbon forms single bonds, it is usually in either sp, sp2, or sp3 hybridization. The single bond is a sigma bond. If the carbon is either sp or sp2 hybridized, then it will have one (in the case of sp2) or two (in the case of sp) unhybridized p-orbitals. The extra p orbitals will form double b...

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He said that there is a newer method, but we will be using the older one.

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The notation is AX_E_, with the _ indicating the number of terminal atoms (X) or number of lone pairs (E). If there are no lone pairs, simply omit the E (you will end up with AX_). If there is one lone pair, omit the subscript (you will end up with AX_E). If there is more than one lone pair, you wil...

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1. Draw VSEPR structures. 2. Draw dipole moments. 3. CO2 has dipole moments, but they cancel out because they are equal in magnitude and opposite in direction. Therefore, CO2 is a nonpolar molecule despite having polar bonds. H2O has dipole moments, but they do not cancel out because they are equal ...

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chromium and copper because they only need one electron to fill half or all of the d orbitals

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If the molecule is not symmetric, then it is (generally) polar because the dipole moments (if any) do not cancel out. If the molecule is symmetric but the outside atoms are not the same, then it is polar because the dipole moments do not cancel out. If the molecule is symmetric and the outside atoms...

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In general, the elements to top right of the periodic table have higher ionization energies. Conversely, the elements to bottom left have lower ionization energies. The elements given in the problem all follow this trend. Just rank them by which ones are higher or righter on the periodic table.

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For HW problem 2.55 c, we are asked to write the valence-shell configuration for the Group 5 transition metals. I got (n-1)d3 ns2, but the answer key says (n-1)d5 ns2. Does the question mean to ask for the fifth group of transition metals or is there some special property of Group 5 metals that make...

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No two electrons in a single atom can have the same four quantum numbers.

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It depends on what the question is asking for. I would use noble gas configuration when the question asks for it.

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It can refer to either. The Angstrom is a unit of length.

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No, it is not necessary.

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I will usually convert the nm into m first because most ratios use m (such as speed of light). After I finish solving the problem, I make sure that the answer is in the units that the question asks for. If the question does not satisfy, I will usually just leave it in meters.

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On number 1.3, the question asks for what happens when the frequency of electromagnetic radiation decreases. The solutions manual says that the correct answer is "The extent of the change in the electrical field at a given point decreases". Why is this the correct answer?

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For net ionic equations, the soluble compounds will dissolve in water. Then, if the separated ions are able to form insoluble compounds, we will have a precipitate. In this problem, the precipitate is Cu(OH)2. The only molecules represented in the net ionic equations are precipitates and the aqueous...

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There are 3 sig figs. The 0's to the left of nonzero numbers do not count as sig figs. However, if there were 0's to the right of nonzero numbers, those 0's would be significant.

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