CHEMISTRY COMMUNITY

Jessica Booth 2F

Comparing pKa/pKb with pH/pOH will be important to know how to do on the final - I think! See...pKa = -log(Ka), so the higher the pKa, the lower the Ka. Ka = \frac{[A-][H+]}{[HA]} Lower Ka means higher products - and lower H+ pH = -log(H+), which means the higher the pH, the lower the H+ Therefore,...

Jessica Booth 2F

It depends on the situation that is being set up. If the cell is a galvanic cell then the half-reaction with the more negative potential will be oxidized and the half-reaction with the more positive potential will be reduced. If the set-up is for electrolysis, then the more negative potential will b...

Jessica Booth 2F

The pre-equilibrium method is used to eliminate an intermediate in the rate law. In the pre-equilibrium method, the step before the slow step can be assumed to bottle-neck, creating a buildup of products. This allows products to go back to reactants and the reaction to be at equilibrium. From then y...

Jessica Booth 2F

The slow step is the step with the highest activation energy. With regard to solving problems we use the slow step to determine if a given mechanism is correct for a reaction and to determine the order and rate law of a reaction.

Jessica Booth 2F

You can tell because of the units of k, but you can also know that it is a first-order reaction because it tells you in the problem.

Jessica Booth 2F

Since the slow step is the rate-determining step, if the catalyst doesn't accelerate the slow step, then the rate of the reaction won't change.

Jessica Booth 2F

At T=0 K, the entropy that a molecule has is just due to the different microstates that the molecule could be in. At t=0 there are no other forms of entropy so you just have to calculate the entropy that arises from the molecule occupying different positions.

Jessica Booth 2F

If there is a graph, you can look at the difference between the top of the bump to the starting energy level of the different fast steps to determine which fast step has a higher activation energy. In other words, the fast step with the largest "uphill section" will have the higher activat...

Jessica Booth 2F

You can solve for k using this version of the Arrhenius equation: $ln k = \frac{-E_{a}}{RT} + lnA$ . To calculate k you need another k value for a different temperature and then you can subtract $lnk_{2}$ from $lnk_{1}$ and the lnA term will cancel.

Jessica Booth 2F

Correct, they should be replaced using the equilibrium equation solved for the intermediate.

Jessica Booth 2F

Yes, the rate-determining step will always have the highest activation energy because the step with the largest activation energy is the slowest step so it is the rate-determining step. To find the rate determining step when looking at the reaction profile, find the step with the largest difference ...

Jessica Booth 2F

Yes, it is because you have to account for the fact that the rate = A^{2} . From this equation, you can get the half-life equation which is t_{1/2}= \frac{1}{k[A]_{0}} . From this equation you can clearly see that with respect to [A], the half-life has an inverse relationship, so the plot won't be l...

Jessica Booth 2F

How do you know if in a multistep reaction profile each starting energy for each step increases or decreases? For example in question 7.11, why doesn't the energy decrease each step instead of increasing?

Jessica Booth 2F

the constant in the pre equilibrium approach is dependent on the coefficients of the species that you are using to determine the rate of. In class, this was the 2 NO. Also the constant depends on the rate constant of the slow step as well as the equilibrium expression that you solve for to replace t...

Jessica Booth 2F

You can use the Arrhenius equation to find the activation energy as well as the rate constant at different temperatures.

Jessica Booth 2F

The equilibrium constant is the forward reaction constant over the reverse rate constant. For part b) you can think of it in terms of le chatlier's principle. Since the equilibrium constant is less than 1 you know that the reactant are favored and the reaction is endothermic. When you raise the temp...

Jessica Booth 2F

For a redox reaction in an acidic half-reaction, do we use H3O+ or just H+ to balance the hydrogens?

Jessica Booth 2F

Galvanic and Voltaic cells are the same. They both get their energy from spontaneous reactions. The E of the cell will be positive and the $\Delta G$ is negative.

Jessica Booth 2F

The reducing agent ends up becoming oxidized because it donates its electron to reduce another species. The oxidizing agent becomes reduced because it gains an electron from another species that becomes oxidized.

Jessica Booth 2F

Cd + 2Ni(OH)3 $\rightarrow$ Cd(OH), why is there KOH on the anode side and Ni on the cathode side?

Jessica Booth 2F

Is there a convention for which order you write the states in when there is an inert electrode. For example, for CL2 + H2 $\rightarrow$ HCl, is there any order in which you have to write the H2 and H+ on the anode side?

Jessica Booth 2F

For $Ce^{4+} + I^{-}\rightarrow I_{2} + Ce^{3+}$ , how can you tell that the are in the same solution and therefore need a inert electrode?

Jessica Booth 2F

Will someone explain how to find the half-reactions for $P_{4} \rightarrow H_{2}PO_{2} + PH_{3}$ ?

Jessica Booth 2F

How do you know that O3 is oxidized? $O3 + Br- \rightarrow O2 + BrO3$

Jessica Booth 2F

Will someone explain how to do part d? Cl2 $\rightarrow HClO + Cl2$

Jessica Booth 2F

You can just keep the moles as grams in the calculations. However, you will need to convert to moles to find the heat to melt the ice as the $\Delta H$ is given in KJ/mol.

Jessica Booth 2F

Because energy isn't lost you know that q1+q2=0 which means that q1=-q2. q1 is the heat required to first melt the ice and then increase its temperature to the final temp. So q_{1}=n\Delta H_{fus} +mC\Delta T . Also, q2 is the heat lost by the water when it is cooled down. So, q_{2}=mC\Delta T . Rem...

Jessica Booth 2F

When do you use the constant R=8.314 vs 8.206x10^-2 and what are the units for each
The units are R = 8.314 J/(K mol) and 8.206x10^-2 L atm/(K mol). You use 8.314 whenever you are working with J and you use 8.206x10^-2 when using atm and L.

Jessica Booth 2F

You should be able to use q=n $\Delta H$ . For part a) $\Delta H$ = 4.76 KJ/ .579 mol = 8.22 KJ/mol. Part b is the same except you need to convert the mass given into moles using the molar mass.

Jessica Booth 2F

The correct kA value was 1.9 x10^-5.

Jessica Booth 2F

After finding the Ka set up your ice table. You should get that the equilibrium conc. are [HClO2]= .15-x, [ClO2-]=x and [H3O+]=x. Then set up your equilibrium equation with the appropriate values. The Kc should = x^2/(.15-x) which also equals the Ka that is .012. Then solve the quadratic and you get...

Jessica Booth 2F

The correct answer for part a is 343 K and the correct answer for part b is 373K

Jessica Booth 2F

First, U=q+w. So you know that q= 5.50KJ. Then you know that there is a constant pressure so $w= -p\Delta V$ .

Jessica Booth 2F

By definition, the enthalpy is the same as the heat transfer under constant pressure. Under conditions of constant volume, there is no work done, so the change in internal energy equals the enthalpy.

Jessica Booth 2F

Adiabatic mean that $\Delta Q = 0$ , so a reaction can still absorb or release energy if work is done on the system or the system does work.

Jessica Booth 2F

Because you are calculating the work done by the system, which is equal to the negative work done on the system. If a gas does work on its surrounding then the final volume is greater than the initial volume so in order to convey that the system lost energy the work must equal $-p\Delta V$

Jessica Booth 2F

Non-expansion work is work where the volume doesn't change. This could be muscle contractions or transmission of nerve impulses. expansion work is work that occurs when there is a change in volume. Work is for the most part independent of enthalpy except for when heat is added or taken away from a s...

Jessica Booth 2F

Yes, there are scenarios where a closed reaction would do work on the system but still have positive enthalpy because the sign of the enthalpy doesn't indicate anything about the work. It is possible for a system to lose heat, having a negative q but do work on the surroundings by expanding. Exother...

Jessica Booth 2F

It does not matter if the products and reactants are in different states. However, you do have to differentiate between the states because there can be different enthalpies of formation for different phases of the same molecule.

Jessica Booth 2F

Bond enthalpies are the energy required to break a bond, while standard enthalpies of formation are the reaction enthalpies for the formation of 1 mol of a substance from its elements in their most table form.

Jessica Booth 2F

Basically you take the sum of the enthalpies of formation of the products and subtract the sum of enthalpies of formation of the reactants.

Jessica Booth 2F

The concentration of H30+ is initially .15 M for the second protonation because .15 M H2SO4 produces .15 M H3O+ so when the second protonation starts there is already .15M H3O+ in solution from the first protonation.

Jessica Booth 2F

Partially. The percent ionization is the amount of ionized product divided by the amount of reactant. So it tells us the ratio of ionized product to unionized reactant.

Jessica Booth 2F

The question is: Calculate the pH of each of the following solutions of diprotic acids at 25 $^{\circ}C$ ignoring second deprotonations only when the approximation is justified. How do you tell if ignoring the second deprotonation is justified?

Jessica Booth 2F

What information were we supposed to find in table 5E. 2? My book didn't have a table 5E. 2.

Jessica Booth 2F

The following plot shows how the partial pressures of reactant and products vary with time for the decomposition of compound A into compounds B and C. All three compounds are gases. Use this plot to do the following: write the balanced chemical equation for the reaction. b) calculate the equilibrium...

Jessica Booth 2F

At 25 $^{\circ}C$ K = 3.2 x $10^{-34}$ for the reaction 2HCl $\rightleftharpoons$ H2 + Cl2. If a reaction vessel of volumme 1.0L is filled with HCl at .22 bar, what are the equilibrium partial pressures of HCl, H2, and Cl2.

Jessica Booth 2F

The general rule for change in pressure is that if the pressure has increased the side of the reaction with fewer moles is favored. This, however, only applies if the pressure is changed by changing the volume because then the concentrations change.

Jessica Booth 2F

It applies to changes in concentration, pressure, or temperature.

Jessica Booth 2F

The ice tables are mainly used when we only know the initial concentration(s) and the equilibrium constant. Basically, we use ice tables when we are unable to just use the equilibrium concentrations and equilibrium constant to find the other equilibrium concentrations.

Jessica Booth 2F

Ka is the acid dissociation constant and is equal to [H+][A-]/[HA]. The pKa is the -log of the Ka. The Ka and pKa tell you the strength of a weak acid. The larger the Ka and the smaller the pKa the stronger the acid.

Jessica Booth 2F

I'm not sure if these are all the ones we need to know but here are some: strong acids: HCl, HI, HBr, H2SO4, HNO3, HClO4. Strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2.

Jessica Booth 2F

No, they would not cancel. The only way to have a nonpolar tetrahedral molecule is if all the atoms attached to the central atom are the same.

Jessica Booth 2F

Are the guidelines for determining how strong a base is the same as the guidelines for relative acidity?

Jessica Booth 2F

Pi bonds are formed using unhybridized orbitals. If a bond is using sp2 orbitals then the pi bonds are formed using p orbitals.

Jessica Booth 2F

'm still not exactly sure what a conjugate acid is. If a molecule gives off a proton in a reaction, in the reverse reaction is it then the conjugate base since it receives a proton? A conjugate acid is a compound that starts out as a base and then gains an H+. If a molecule gives off an H+ in a reac...

Jessica Booth 2F

To calculate the pH of a solution take the -log of the concentration of H3O+ ions.

Jessica Booth 2F

Amphiprotic means that the compound can act as both an acid and a base by either accepting or donating an H+. The easiest way to answer this question is to write the chemical reaction between the given compound and H2O. Then show the products when the compound is acting as an acid and as a base. For...

Jessica Booth 2F

Why is HCOOH a stronger acid than CH3COOH? I would assume that CH3COOH would be stronger because of the larger delocalization of charge caused by the CH3 group?

Jessica Booth 2F

part c asks whether HBrO2 or HClO2 is a stronger acid. the H-Br bond is longer and therefore weaker than the H-Cl bond, so why is HClO2 the stronger acid?

Jessica Booth 2F

F is more electronegative than Cl so it is less likely to give off the H+ ion. This means that HCl can disassociate easier, making it the stronger acid.

Jessica Booth 2F

You can count the number of regions of high electron density (a bond or lone pair) and then from there assign it a hybridization with the sample number of orbitals as electron densities. For example if the molecule had 3 areas of high electron density then you would choose a hybridization that has 3...

Jessica Booth 2F

To find the coordination number count the number of bonds that the transition metal forms. For a) the Ni is bonded to 4 Cl so the coordination number is 4. Remember that en is bidentate and edta is hexadentate.

Jessica Booth 2F

It has a coordination number of 6 because it forms a bond with each of the Cl's and then each of the en is bidentate so each en bonds twice to the Pt and since there's 2 of them that's 4 bonds. Then the total number of bonds the Pt has is 6.

Jessica Booth 2F

Cheating ligand form a ring of atoms that include the central metal atom. Chelating ligands have to be multidentate the chelate.

Jessica Booth 2F

First, you count up all the valence electrons, 2(5) + 6= 16. Then you put the less electronegative atom in the middle, which is N. Then you arrange the O and other N symmetrically around the central O. Then you put the 16 electrons around the atoms. You find out that with just single bonds, not all ...

Jessica Booth 2F

What is the difference between an ion-ion molecular force and an ionic bond or are they the same thing?

Jessica Booth 2F

Terminal atoms are just the atoms surrounding the central atom in a Lewis Structure. For example, Cl is a terminal atom in CCl4.

Jessica Booth 2F

Correct. We will not be asked to draw them but we will need to name the shapes.

Jessica Booth 2F

Yes, seesaw and sawhorse both have the VSEPR model AX4E

Jessica Booth 2F

Use Lewis Structures and the VSEPR model to give the VSEPR formula for each of the following species and predict its shape:
a)sulfur tetrachloride
b)iodine trichloride
c) IF4-
d)xenon trioxide
What is the VSEPR formula for each part and how do you find it?

Jessica Booth 2F

It experiences a dipole moment because the Cl's and H's are next to each other and so there is a dipole moment pointing from the H to the C and from the C to the Cl. So, therefore, the dipole moments add in both directions so the dipole moment points from the H to the Cl.

Jessica Booth 2F

The bigger the atom, the more polarizable it will be because the outer electrons feel the pull of the nucleus less so they can be more distorted. The number of electrons is indicative of the size.

Jessica Booth 2F

How do you determine that in the lewis structure of CH2Cl2 the Cl's are next to each other and the H's are next to each other instead of the Cl's and H's being across from each other?

Jessica Booth 2F

Why does H2SeO4 have a dipole moment?

Jessica Booth 2F

Electron affinity is the energy released when an electron is added to a gas-phase atom. Electronegativity is the electron-pulling power of an atom when it is part of a molecule. Electronegativity is the average of the ionization energy and electron affinity of an element.

Jessica Booth 2F

Yes, all 3 of the 2p orbitals and all 5 of the 3d orbitals are degenerate because they are at the same energy level.

Jessica Booth 2F

Polarizability is how much an electron cloud of an atom or ion can undergo a large distortion. Atoms or ions with a large distortion are highly polarizable. The larger the atom the more polarizable it is.

Jessica Booth 2F

"Use the covalent radii in Fig. 2D.11 to calculate the bond lengths in the following molecules. Account for the trends in your calculated values: (a) CF 4; (b) SiF 4; (c) SnF 4" In order to find the bond length of covalent bonds you need to add together the 2 covalent radii for the element...

Jessica Booth 2F

Draw the complete Lewis structure for each of the following compounds:
c) H2C(NH2)COOH
How do you determine the arrangement of the atoms and which atoms are bonded to another?

Jessica Booth 2F

This is the question: Which M2+ ions (where M is a metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d7; (b) [Ar]3d6; (c) [Kr]4d4; (d) [Kr]4d3? a) Because there is a 2+ charge we need to add 2 electrons to the given number of valence electrons, which in this c...

Jessica Booth 2F

You start from the left side of the periodic table and count across a period until you reach the element you're trying to find. The count is the number of valence electrons. If you're trying to find the number of valence electrons for a cation then you find the number of valence electrons and subtra...

Jessica Booth 2F

C,N,O,F always form an octect and anything with an atomic number above F has a d-shell, so they can form expanded octects.

Jessica Booth 2F

Cations are smaller than their parent atoms because there is less electron-electron repulsion but there is still the same number of protons so the electrons are closer to the nucleus.

Jessica Booth 2F

Sb and Mn have electrons in the d-shell so those electrons are counted as valence electrons because they are in the outermost shell.

Jessica Booth 2F

Electrons are removed from the highest energy orbital and the energies of orbitals go s<p<d<f.

Jessica Booth 2F

1E.25 Give the notation for the valence-shell configuration
(including the outermost d-electrons) of (a) the alkali metals;
(b) Group 15 elements; (c) Group 5 transition metals; (d) the
“coinage” metals (Cu, Ag, Au).
a) ns^1
b) ns^2 np^3
c) omit
d) (n-1)d^10 ns^1

Jessica Booth 2F

Electrons in an s-orbital are more effective at shielding other electrons because the penetration levels go ns > np > nd > nf. The higher the penetration the more effective an electron is at shielding others. The s-orbital is closest to the nucleus so it is the most effective at shielding other elec...

Jessica Booth 2F

what were the electron configuration exceptions, and does anyone know why they are exceptions? additionally, other than asking what the exceptions are on an exam, what kind of questions would test this knowledge of exceptions? The exceptions that we need to know are Cr whose electron configuration i...

Jessica Booth 2F

The s < p < d < f is referring to the energy levels of different subshells. They can all exist at the same time for example boron has electrons in the 2s and 2p orbitals. The only limitation is if the atom doesn't have enough electrons to need to fill the p, d, or f levels. For example, magnesium do...

Jessica Booth 2F

keV is kilo-electron volts, which is a unit for energy. The conversion from kilo- electron volts to electron volts is (1000 eV/keV) and then the conversion from eV to J is $\left ( 1.602 * 10^{-19} \right )$ J/ 1eV.

Jessica Booth 2F

The speed in the question is referring to the speed of light which is always $3.0 \ast 10^{8} m/s$

Jessica Booth 2F

Will someone explain how to find the final and initial energy levels of an electron during the emission of energy that leads to a spectral line observed at 102.6 nm in the ultraviolet spectrum of atomic hydrogen using the equation from class?

Jessica Booth 2F

Yes ,if instead b had said the wavelength increases it would have been correct because c= $c=\lambda \nu$ so wavelength and frequency have an inverse relationship so when one increased the other has to decrease.

Jessica Booth 2F

Why is the speed of the radiation constant when the frequency of EM radiation decreases?

Jessica Booth 2F

Using the fact that n = MV, by rearranging the equation you get V = n/M. 4.50 mmol = 4.5 x 10^-3 m so V = (4.5 x 10^-3 m)/ (.278 M) = 16.2 mL

Jessica Booth 2F

In homework L35 it asks you to use a mass of 2.50 t. What is the conversion from t to grams?

Jessica Booth 2F

Also the last equation that you balanced isn't balanced because the textbook has an error and the last equation should be Fe3Br8 + Na2CO3 forms NaBr + CO2 + Fe3O4.

Jessica Booth 2F

Nevermind, the textbook has an error on problem L35. The last equation should be Fe3Br8 + Na2CO3 forms NaBr + CO2 + Fe3O4.

Jessica Booth 2F

How do you balance this equation?
FeBr2 + Na2CO3 goes to form NaBr + CO2 + Fe3O4

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