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« Reply #3735 on: April 15, 2019, 06:39:15 pm »
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Hey there,
What is the difference between primary, secondary and tertiary alcohols (organic chemistry)? Thanks
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#### Rom_Dog

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« Reply #3736 on: April 15, 2019, 06:55:40 pm »
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Hey!

Structurally there is a clear difference between primary, secondary and tertiary alcohols. The carbon atom attached to the hydroxyl (-OH) group of a primary alcohol is only attached to one carbon atom, in secondary alcohols this carbon is attached to 2 other carbon atoms and in tertiary alcohols this carbon is attached to 3 other carbon atoms.

Physically there are also differences. Primary, secondary and tertiary alcohols with the same molecular formula have different melting and boiling points, this is because their structure has an effect on the intermolecular bonds between alcohol molecules.

There is also a chemical difference. When oxidised primary alcohols form carboxylic acids (with an aldehyde as an intermediate state), secondary alcohols are oxidised into ketones and tertiary alcohols cannot be oxidised!

These are the keys differences that you need to know for VCE chemistry, I'm not sure which of these points are relevant for HSC chem but I hope that this helps!

- Rom
« Last Edit: April 15, 2019, 06:58:54 pm by Rom_Dog »

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#### 006896

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« Reply #3737 on: April 23, 2019, 09:35:25 am »
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Hi,
Could you help confirm the following statements, regarding equilibrium reactions.
If adding an inert gas to a system while keeping the volume the same, no change occurs.
- How would you keep the volume the same while adding an inert gas? Does it mean the volume of the reacting gases, or the volume of the system?
If adding an inert gas to a system while keeping the pressure the same, the volume must increase and the equilibrium shifts to the side with more gas volume.
- I don't really understand this statement.

Thanks

#### emmajb37

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« Reply #3738 on: April 23, 2019, 12:30:52 pm »
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Hey, I was just doing Question 7, Section2 of the Solution Equilibria Test 1 in the new chemistry topic test book.
I am really confused because the answers say that potassium chloride is soluble because of the PMS. But I thought the P was for lead.
As well as that potassium is a group 1 ion which is always soluble. Just really confused and would love some clarification.
Thanks,
Emma

#### emmajb37

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« Reply #3739 on: April 23, 2019, 12:33:38 pm »
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Hey, I was just doing Question 7, Section2 of the Solution Equilibria Test 1 in the new chemistry topic test book.
I am really confused because the answers say that potassium chloride is soluble because of the PMS. But I thought the P was for lead.
As well as that potassium is a group 1 ion which is always soluble. Just really confused and would love some clarification.
Thanks,
Emma
And also this is just part a) and the rest of the question parts b) and c) continue to use KCl as a solid.

#### iktimal

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« Reply #3740 on: April 24, 2019, 09:50:39 pm »
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Hey, Can someone please explain to me how to do part D.
Ksp= 1.41 x 10^-12
Thanks!

#### myopic_owl22

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« Reply #3741 on: April 25, 2019, 06:06:14 pm »
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Hi,
Could you help confirm the following statements, regarding equilibrium reactions.
If adding an inert gas to a system while keeping the volume the same, no change occurs.
- How would you keep the volume the same while adding an inert gas? Does it mean the volume of the reacting gases, or the volume of the system?
If adding an inert gas to a system while keeping the pressure the same, the volume must increase and the equilibrium shifts to the side with more gas volume.
- I don't really understand this statement.

Thanks

Hey there,
I know it's a little old, but this thread explains the issue of inert gases much better than I ever could (it's to do with the gas not affecting the partial pressures of the reagent gases): https://atarnotes.com/forum/index.php?topic=7350.0

1) is correct. Adding the inert gas won't change the number of moles of the reagents of the vessel, and therefore nothing happens to the equilibrium. Assuming the reaction vessel is rigid, there won't be any change in volume anyway. If you meant to say pressure, then yes, the total pressure of the system does change, but we're looking at partial pressures so it's irrelevant.

2) From the stackexchange website, the equilibrium position will change if the volume changes when adding an inert gas (which will happen if the pressure is kept constant - Boyle's law) - in the specific case where the reaction vessel is not rigid - e.g. a balloon. Increasing the volume of the system will affect the partial pressures of the reagents - they will decrease as the inert gas is added (as there's more volume for the same moles of reagent) and thus the equilibrium will shift to increase the pressure of the vessel by increasing the moles it contains.

Hopefully this was a little useful! Might I add, this whole concept is really weird - it's one of those things you just accept and move on, I guess
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#### myopic_owl22

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« Reply #3742 on: April 25, 2019, 06:40:56 pm »
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Hey, Can someone please explain to me how to do part D.
Ksp= 1.41 x 10^-12
Thanks!

Hey there,
There are a few steps involved, but hopefully I'm making some sense!
1. Write an equation (as always)
$Mg(OH)_{2(s)} \rightleftharpoons Mg_{(aq)}^{2+} + 2OH_{(aq)}$
Note the molar ratios - the OH has 2.

2. Write a Ksp expression:
$[Mg^{2+}]\cdot [2OH]^{2} = 1.41\times 10^{-12}$
That's the solubility product, considering the OH's molar ratio. We don't divide by the magnesium hydroxide as that's a solid when it isn't dissolved.

3. Replace the concentration of Mg2+ with x. We know this to be the same concentration as that of magnesium hydroxide (what we're trying to figure out).
$x\cdot 4x^{2} = 1.41\times 10^{-12}$

4. Solving for x gives: 7.06 *10-5M. This is the molar solubility of Mg(OH)2 - the amount of moles that can be dissolved in a litre of water.

An aside: regular solubility is measured in g/L. To convert, just multiply 7.06 *10-5 by the molar mass of Mg(OH)2
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#### mani.s_

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« Reply #3743 on: April 26, 2019, 09:59:58 am »
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I just had a question for the chemistry notes, why is Carbon 12 considered a isotope when it is the normal/natural form of carbon, having 6 protons, 6 neutrons and 6 electrons.

Thanks

#### insanipi

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« Reply #3744 on: April 26, 2019, 10:25:18 am »
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I just had a question for the chemistry notes, why is Carbon 12 considered a isotope when it is the normal/natural form of carbon, having 6 protons, 6 neutrons and 6 electrons.

Thanks
Isotopes are forms of the same element- they have the same number of protons, but each isotope has a different number of neutrons. Let's look at Carbon-12 and Carbon-14. They both have 6 protons. However, Carbon-12 has 6 neutrons whilst Carbon-14 has 8 neutrons. By definition this means that Carbon-12 and Carbon-14 are both isotopes of the carbon atom.

Carbon-12 is the most abundant isotope of carbon, whereas Carbon-13 and Carbon-14 are less abundant.
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#### mani.s_

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« Reply #3745 on: April 26, 2019, 11:09:53 am »
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Isotopes are forms of the same element- they have the same number of protons, but each isotope has a different number of neutrons. Let's look at Carbon-12 and Carbon-14. They both have 6 protons. However, Carbon-12 has 6 neutrons whilst Carbon-14 has 8 neutrons. By definition this means that Carbon-12 and Carbon-14 are both isotopes of the carbon atom.

Carbon-12 is the most abundant isotope of carbon, whereas Carbon-13 and Carbon-14 are less abundant.
Does that mean every element is an isotope of that element???

#### myopic_owl22

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« Reply #3746 on: April 27, 2019, 02:12:48 pm »
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Does that mean every element is an isotope of that element???

Sure does
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Also, I've got a question from the NESA sample paper: Can someone explain why the answer would be C?

Thanks a lot!
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#### Bookdragon_202

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« Reply #3747 on: May 08, 2019, 10:26:28 am »
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I have a titration assessment coming up and am feeling a little overwhelmed. What is the best way to study for this assessment? It is a practical with some theory questions.

#### stella_atarnotes

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« Reply #3748 on: May 08, 2019, 06:24:24 pm »
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I have a titration assessment coming up and am feeling a little overwhelmed. What is the best way to study for this assessment? It is a practical with some theory questions.

Hey! With titrations, just understand what solutions you are putting into the burette, pipette, volumetric flask and know the common indicators (e.g. phenolphthalein) and the colour change. Know what to wash your instruments with really well. Also work on some practice titration questions from textbooks or past exams to make sure you know how to write up your results. Also try to know some sources of errors (e.g. washing the burette with water will dilute the solution you put into it) Other than that, just keep calm and trust your knowledge. You'll do fine!

#### stella_atarnotes

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« Reply #3749 on: May 20, 2019, 07:10:10 pm »
+1
Sure does
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Also, I've got a question from the NESA sample paper: Can someone explain why the answer would be C?

(Image removed from quote.)

Thanks a lot!

You can eliminate options A and B because these are acids and have a relatively low pH. With C and D, they have used the acids given to you in the question stem and formed a salt with it. Remember that weak acids and weak bases combine to form a neutral salt, but if you have weak acids combining with strong bases (NaOH in this case), the salt will be slightly alkalinic. Since Nitrous acid already has a higher pKa, the salt formed when it is combined with a strong base will be more alkalinic than a salt formed with chlorous acid. Thus it is C