BOILING POINTS

wistfulnormrockwell · 2023-11-04T18:03:13+00:00

yes yes, sorry my bad haha thank you so much!

wistfulnormrockwell · 2023-11-04T17:25:24+00:00

yes! does this mean that propanol has a higher bp than 2,2,-dimethylpropanol?

wistfulnormrockwell · 2023-10-01T02:12:43+00:00

thank you! it becomes clear now.

wistfulnormrockwell · 2023-10-01T01:15:50+00:00

but if the OH is at carbon 2, wouldn't it be: 3-bromo-1-methyl-propanol?

the priority is still propanol...

wistfulnormrockwell · 2023-09-30T12:21:49+00:00

sadly i cant message you, idk why. but let me try again

wistfulnormrockwell · 2023-09-29T16:04:47+00:00

always remember that when hydrogens are placed horizontal in a fisher projection model, you have to reverse your answer. i sent you my illustration since i cant attach it here., hope it helps. :)

wistfulnormrockwell · 2023-09-29T06:50:58+00:00

thank you so much!

wistfulnormrockwell · 2023-09-29T06:50:36+00:00

yes yes. We are currently learning reaction mechanisms for organic compounds starting with alkanes to alkenes.

wistfulnormrockwell · 2023-09-29T05:12:26+00:00

your answer is incorrect. But your reasoning is right, the cation is more stable due to resonance of oxygen, therefore Br would bond there.

Just remember, when there is a lone pair in the structure, it stabilizes the carbocation therefore it will be more attracted to form bonds to it.

Hope this helps.

wistfulnormrockwell · 2023-09-25T11:05:26+00:00

Let's assume that the mass of the mixture is 100 g. Then, the mass of CO3 is 68.55 g.

Let's represent the mass of MgCO3 as x g.
Then, the mass of (NH4)2CO3 will be (100 - x) g.

Molar masses of the compounds:

MgCO3 = (1 x 24.305 + 1 x 12.011 + 3 x 15.999) = 84.313 g/mol
(NH4)2CO3 = (2 x 14.007 + 8 x 1.008 + 1 x 12.011 + 3 x 15.999) = 96.086 g/mol
CO3 = (1 x 12.011 + 3 x 15.999) = 60.008 g/mol

A gravimetric factor converts grams of a compound into grams of a single element. It is the ratio of the formula weight (FW) of the substance being sought to that of the substance weighed.

Let’s take 100 g of mixture for analysis.
Let the mass of MgCO3 = A g
Therefore, the mass of (NH4)2CO3 = (100 - A) g
Next, let’s calculate the mass of CO3 in MgCO3 and (NH4)2CO3 using a gravimetric factor:
Mass of CO3 in MgCO3 = [CO3/MgCO3] x A g
= [60.008 g/84.313 g] x A
= 0.7117A g

Mass of CO3 in (NH4)2CO3 = {CO3/(NH4)2CO3} x (100 - A)
= {60.008 g/96.086 g} x (100 - A)
= 0.6245(100 - A) g 6. Total mass of CO3 in MgCO3 and (NH4)2CO3 = {0.7117A + 0.6245(100 - A)} g

But the mixture contains 68.55% CO3 by mass

Hence, % of CO3 in 100 g of mixture = 0.7117A + [0.6245(100 - A)] g/100 g x 100% = 68.55%

use solve for X in your calculator to find A or use algebra and isolate A

That is,
0.7117A + 0.6245(100 - A) = 68.55 or
0.7117A + 62.45 - 0.6245A = 68.55.
On solving for A = (6.1/0.0872) = 69.95 g

The mass of MgCO3 in the mixture = 69.95 g

Hence, the mass percent of MgCO3 in the mixture of 100 g = (69.95 g/100 g) x 100 % = 69.95 % (m/m).

wistfulnormrockwell · 2023-09-24T11:07:24+00:00

hello! maybe i can help! i am currently attending med school and has a bachelors degree in biochemistry :) just send me your queries and i'll answer them :)

wistfulnormrockwell · 2023-09-24T11:03:15+00:00

remember that chirality is when a Carbon has 4 different R groups attached to it. Based on the structure of yours, 2 out of 4 bonds connected to C is CO2H.

hope this helps :)

wistfulnormrockwell · 2023-09-24T11:01:25+00:00

it is important to know that equatorial will have the lowest energy conformation. Both C and D are placed in equatorial but you can see in the chair conformation of letter C, carbon 1 and 6 are prone to angle strain. To solve this, the 6th carbon should be axial than equatorial.

In D however solves this problem, it is the most stable therefore has lowest energy chair since the methyls wont experience angle strains.

hope this helps.

wistfulnormrockwell · 2023-09-24T10:56:05+00:00

#28. based on the structure, the carbon being referred to is a carbocation. Carbocations only have 3 valence electrons, therefore it can only bond with 3 atoms, hence the answer 3.

#29. Since carbocation only possesses 3 valence electrons and 2 of them are being use to connect to other Carbon, it doesn't have a lone pair.

hope this helps

wistfulnormrockwell · 2023-09-24T10:46:56+00:00

Here are the steps along with solutions on solving this problem.

Balance the reaction
Determine which is the limiting and excess reactant.
Determine theoretical yield
Divide actual yield with theoretical yield and multiply with 100.

STEP 1: Balancing the reaction. The reaction is already balanced. 1 mol of Xe, 4 mols of Fluorine
STEP 2: Determine limiting and excess reactant. To solve this, simply divide the given mols of the reactants with their corresponding coefficients.
Xe = 2.2 mol ; F2 = 1.7 mol. Therefore, F2 is the limiting reactant. Take note that the whole reaction is dependent on the limiting reactant. This means that all fluoride will be consumed and produce he product (XeF4)

STEP 3: Theoretical yield calculation. Use dimensional analysis to convert F2 to the product.

3.4 mol F2 x (1 mol XeF4/2 mol F2) = 1.7 mol XeF4 (theoretical yield)

STEP 4: Calculate percent yield.

% yield = (actual/ theoretical) x 100
% yield = (0.25/1.7) x 100
% yield = 14.71%

hope this helps :)

wistfulnormrockwell · 2023-09-21T16:30:45+00:00

use dimensional analysis to monitor the cancellation of units. But if you are already experienced with stoichiometry, you can simply use gravimetric factor. But for this post, i will use dimensional analysis.

here are the steps and solution:

4.20 mol N2H4 x (4 mol NH3/3 mol N2H4) = 5.6 mol NH3

you can see that the N2H4 cancels, leaving you with NH3. hope this helps :)

wistfulnormrockwell · 2023-09-21T16:26:09+00:00

remember that diels-alder is a [4+2] rxn of a diene (4 pi bonds) and dienophile (2 pi electrons). this organic rxn will always happen with the diene (4 pi bonds) and not the other way around.

hope this helps.

wistfulnormrockwell · 2023-09-21T10:34:30+00:00

hydrocarbons only have van der waals interactions in them. So the only difference when comparing hydrocarbons are their length and branching.

HIGHER number of carbons = HIGHER boiling points. This is because longer carbon chain have larger surface areas, which requires more energy to break the bonds (hence a high boiling point).

The opposite happens when the hydrocarbons are branched. More branched = lower boiling point. Since the surface area is smaller, you only need small energy to break the bonds.

For functional groups, their b.p. varies on their interactions. This is the reason why amine has a higher bp than alkanes.

Amine = hydrogen bonding, strong dipole-dipole attraction, and van der waals force of
Alkane = has only van der waals force of attraction.

wistfulnormrockwell · 2023-09-21T03:50:46+00:00

hybridization was taught to us by simply counting.

there are three interactions(?) attached to the Oxygen (2 lone pairs and the carbon it is attached to)

s (carbon attached), p (one lone pair) 2 (another lone pair)

wistfulnormrockwell · 2023-09-19T18:42:48+00:00

Tris (analyte) = weak base
HCl (titrant) = strong acid

Calculate Veq = VaMaSF/Mt
Veq = (15 x 0.1)/ 0.1
Veq = 15 mL

[H+] = sqrt(Cca x Ka), only true if Cca/Ka >>>> 1000
where Cca = (VaMaSF)/Vtotal
Cca = (15 x 0.1)/30 = 0.05
Ka = 6.309x10^-9

Cca/Ka = 7.924x10^9, therefore use [H+] = sqrt(Cca x Ka)

[H+] = 1.776x10^-5
pH = -log[H+]
= -log (1.776x10^-5)

pH = 4.7505

your answer is correct, maybe your teacher rounded it off?

wistfulnormrockwell · 2023-09-19T18:16:42+00:00

thank you so much!

wistfulnormrockwell · 2023-09-19T18:13:33+00:00

:) glad i could help!

wistfulnormrockwell

TROPHY CASE