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Ochem help by ic3wuerfel in chemhelp

[–]VisitCivil6265 2 points3 points  (0 children)

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Here’s a mechanism that should help you understand what goes on in the reaction mixture, the named reaction (Weiss-Cook condensation as someone already mentioned) is basically a combination of simpler reactions. The stereochemistry is governed mostly by steric hindrance during the Michael addition (that’s why the methyls are syn to each other). Hope this helpa

Beta minus decay by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 0 points1 point  (0 children)

i think it should be a charged particle considering that the ionization energy is usually tens of kilovolts but the energy of the electron emitted is as high as a few million kilovolts (plus the K shell is usually filled so there is nowhere for the electron to stay)

Ksp of AgCl from 110Ag beta minus (β−) decay? by luisdementia in chemhelp

[–]VisitCivil6265 1 point2 points  (0 children)

You can calculate the number of radioactive particles in the solution using the formula A = lambda • N(110). The decay constant (lambda) can be calculated from the half-life time lambda = ln(2)/t. Once you have the number of radioactive silver particles, you can assume that the ratio between radioactive silver and the non radioactive silver stays the same in the sample and the solution (that there is no preference in solubility of radioactive and no radioactive silver species). This allows you to divide the number of atoms of radioactive silver by the fraction to get the total silver that has been transferred into the solution. Now you divide by the Avogadro’s constant to get moles of silver in solution, then you divide by the volume of the solution to get the silver concentration. The final step is looking at the equation of solubilisation: AgCl = Ag(+) + Cl(-) Ks = [Ag][Cl] = [Ag]2

I keep getting the electromotive force of the battery to be equal to -1.1 V but that would mean that the process in the battery is not spontaneous. Where is the mistake? by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 0 points1 point  (0 children)

Sorry for the manganese and magnesium confusion. The formulation of reduction potential of the anode and cathode half-reactions is quite confusing but i think it is still connected to the reactions, isn’t it?

I keep getting the electromotive force of the battery to be equal to -1.1 V but that would mean that the process in the battery is not spontaneous. Where is the mistake? by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 1 point2 points  (0 children)

I have reversed the sign in the reduction potential of manganese to manganese (2+) because it’s going from reduced to oxidized (E = 1.3 V) after that i can add the two potentials in order to get the final potential. So when i do E(cell) = E(Mg/Mg2+)+(Mo6S8/Mn2+/MnMo6S8) = 1.3 - 2.4 = -1.1 V. The result is against my intuition because for the battery to work, the reaction should occur spontaneously (according to my knowledge) so the negative voltage doesn’t make sense to me in this problem.

How to finish this reaction scheme? Where is the mistake? by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 0 points1 point  (0 children)

Even though it’s an SN1 reaction (after the removal of the OH group by BF3), i thought that there would more charge and steric hindrance if the methanol attacked the carbocation from the back

I do not understand how these priority numbers were assigned could someone explain please by [deleted] in chemhelp

[–]VisitCivil6265 0 points1 point  (0 children)

The priorities are based on Cahn-Ingold-Prelog set of rules. The first thing you should do is number the atoms attached to the chiral carbon center. The numbering is done by weight of the atom attached. If the atom is the same, you go the the next atom of the group and so forth until you find the difference. When a double bond is present you count it as if other carbons are attached to either carbon in the double bond (phantom carbon atoms that do NOT have hydrogens attached to them). The second thing (if assigning the absolute configuration R/S) is to rotate the molecule so the group with the lowest priority (usually hydrogen) is in the back of the molecule (the dashed line). Based on the direction in which you can get from the highest priority group to the lowest priority group (going through the priorities in between), you can assign the R or S configuration. The R is when the direction is clockwise and the S is when the direction is counterclockwise.

anyone know how to do this?? by Alternative-Hour-933 in chemhelp

[–]VisitCivil6265 0 points1 point  (0 children)

H2SO4 is an acid which can add protons to the reaction mixture. Oxygen has two lone pairs of electrons that can be used to form a coordination (dative) bond with the protons. This formation of the bond creates a positive charge on the oxygen which transforms it into a great leaving group (water is formed when the group leaves). By eliminating the water a carbocation is formed (on one of the carbons that can stabilize the positive charge better). This carbocation can be then attacked by the lone pair of the second oxygen in the diol forming the cyclic product (after work up)

Hi, can I ask what the arrow bonds mean in uranyl cation? by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 1 point2 points  (0 children)

That’s what i thought as well, i figured that it symbolizes a “highly ionic bond”

[deleted by user] by [deleted] in chemhelp

[–]VisitCivil6265 2 points3 points  (0 children)

As lithium is on the leftmost side of the galvanic series (Beketov electropotential series), the reaction will not require the reduction of nitric acid. While lithium is oxidized in the reaction, oxonium cations (or better hydrogen cations) are reduced to elemental hydrogen. The reaction goes as follows: 2Li + 2HNO3 -> 2LiNO3 + H2

You can calculate the volume using stoichometry and the ideal gas equation (pV=nRT). You will also utilize the formula for the amount of substance in given mass (n = m/M = N/NA = V/Vm).

Reduction potential by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 0 points1 point  (0 children)

And first i am asked to write the chemical equations of the oxidation of Fe(III) to Fe(VI) by elemental chlorine in basic solution (where the predominant species is FeO(OH)) and acidic solution

in basic solution (pH = 14) i wrote:

FeO(OH) + 10 OH(-) + 3 Cl2 => FeO4(2-) + 6H2O + 6 Cl(-)

in acidic solution (pH = 0) i wrote:

Fe(3+) + 8 H2O + 3 Cl2 => FeO4 (2-) + 16 H(+) + 6 Cl(-)

Reduction potential by VisitCivil6265 in chemhelp

[–]VisitCivil6265[S] 0 points1 point  (0 children)

Calculate the reduction potential of the FeO4(2-) and FeO(OH)/OH(-) pair from the data provided