VCE Chemistry Question Thread

[ekay]

Because pentan-2-amine has no other functional group that takes precedence over the amino group (e.g. carboxyl, hydroxyl), this is the preferred name. In your second example, that would indeed be correct because now you've got a carboxyl which takes precedence, hence 2-aminopentanoic acid (and NOT pentan-2-aminoic())

ahh i see ok thanks a lot grannysmith!

[althepal]

Hi everyone.

Just wondering with the molecule 2,2,3-trimethylbutane, are the numbers at the start necessary? Or would it be trimethylbutane since any way a butane molecule with three methyl groups attached is arranged, it is the same molecule.

Thanks!

[alchemy]

Hi everyone.

Just wondering with the molecule 2,2,3-trimethylbutane, are the numbers at the start necessary? Or would it be trimethylbutane since any way a butane molecule with three methyl groups attached is arranged, it is the same molecule.

Thanks!

it's reduntant

[I am a unicorn]

Hello

Why do endothermic reactions require more activation energy than their corresponding exothermic reaction??
i.e. A+B --> C requires more activation energy than C --> A+B   (why is this so?)

Thanks!

[lzxnl]

Hello

Why do endothermic reactions require more activation energy than their corresponding exothermic reaction??
i.e. A+B --> C requires more activation energy than C --> A+B   (why is this so?)

Thanks!

Draw out one of those camel-hump diagrams with the energy and the reactants

[jyodesh.com]

Hello

Why do endothermic reactions require more activation energy than their corresponding exothermic reaction??
i.e. A+B --> C requires more activation energy than C --> A+B   (why is this so?)

Thanks!

If you compare the reaction profiles of exothermic and endothermic reactions (attached) you can see that the activation energy (Ea) of the endothermic reaction is higher than the exothermic reaction. In going from A+B --> C, the reactants must reach the same transition state (the peak) as the reaction from C --> A+B. The difference between an endothermic and exothermic reaction is whether the product has more energy stored in their bonds (endothermic) or less energy stored in their bonds (exothermic).Exothermic reaction release energy from its chemical bonds thus dH=-ve whilst endothermic reactions result in an increase in the energy of chemical bonds thus dH=+ve. Because the bonds in A+B start in a lower energy state than the bonds in C, the reaction between A+B (endothermic) requires a higher activation energy to reach the transition state than the reaction of C (exothermic).
A quick look at the reaction profiles shows that the activation energy of the endothermic reaction (A+B --> C) is equal to the activation energy of the reverse exothermic reaction plus the change in enthalpy  of the exothermic reaction (C--> A+B)

[SwagG]

A 20.00 ml solution containing Ca2+ at a concentration of 20 ppm is to be diluted to a
0.0025 M solution. The volume of water that needs to be added to the 20.00 ml solution is
closest to
A 20 ml
B 40 ml
C 50 ml
D 100 ml

I was thinking you approach the question using c1v1=c2v2, but I'm not sure how to get ppm into M, or am I attempting the question in the wrong manner. I know ppm= mg/L but I am confused haha

[jyodesh.com]

A 20.00 ml solution containing Ca2+ at a concentration of 20 ppm is to be diluted to a
0.0025 M solution. The volume of water that needs to be added to the 20.00 ml solution is
closest to
A 20 ml
B 40 ml
C 50 ml
D 100 ml

I was thinking you approach the question using c1v1=c2v2, but I'm not sure how to get ppm into M, or am I attempting the question in the wrong manner. I know ppm= mg/L but I am confused haha

20ppm=20mg/L

First step is to convert mg/L to M using c(mol/L)=c(g/L)/M(g/mol)
20x10^-3g/L
               
40.1g/mol      = 4.99x10^-4M=c1

since 0.0025M is more concetrated than 0.0005M im kinda thinking that when typing this out you might have missed a zero somewhere?

anyway,

V1=20.00mL
and because we're adding water to calcium solution:
V2=V1+V3, where V3 is the volume of water we add and the value we need to work out.
so this means:
c1V1=c2(V1+V3)
c1V1=c2V1+c2V3
c1V1-c2V1=c2V3
V1(c1-c2)=c2V3
V1(c1-c2)/c2=V3
try and make this derivation yourself, plug in the values and see what you get

[Sundal]

So my textbook says that different analytical techniques are based on the physical and/or chemical properties of the substance being tested.

And then chromatography has been said as being a technique based on both physical and chemical properties of a substance.
How is this so?

[sunshine98]

So my textbook says that different analytical techniques are based on the physical and/or chemical properties of the substance being tested.

And then chromatography has been said as being a technique based on both physical and chemical properties of a substance.
How is this so?

According to my textbook chromatography is
-Physical because the compound is merely being separated not altered.
-Chemical because weak bonds are formed when the molecules adsorb to a surface.

[Sundal]

Do inorganic catalysts also have active sites?

Thanks

[mahler004]

Do inorganic catalysts also have active sites?

Thanks

Kind of. An active site is conventionally just used to refer to an enzyme's active site, though.

[BakedDwarf]

Does a weak acid form a strong conjugate base?
Does a weak base form a strong conjugate acid?

[lzxnl]

Does a weak acid form a strong conjugate base?
Does a weak base form a strong conjugate acid?

No. Weak acids have weak conjugate bases and vice versa.  If an acid has a strong conjugate base, then the base would be fully protonated in solution meaning the original acid would have no acid properties

Weaker acids do have weak conjugate bases that get stronger though as the acid strength decreases

[BakedDwarf]

Why is a pH change in a strong acid greater than a pH change in a weak acid after a dilution?

Say, for example, we have two solutions, 10mL of 0.10M hydrochloric acid and 10mL of 0.10M propanoic acid. They are both diluted to 100mL. However, the pH of HCl before dilution is 1 and after is 2 (it increases by 1). The pH of propanoic acid before dilution in 2.9 and after is 3.4 (increases by 0.5). Can someone explain why this is? I've got a vague idea on why in my head but I just can't put them into words.