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November 01, 2024, 10:55:50 am

Author Topic: VCE Chemistry Question Thread  (Read 2466733 times)  Share 

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Ionic Doc

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Re: VCE Chemistry Question Thread
« Reply #7935 on: March 16, 2019, 05:21:59 pm »
0
   Sodium chloride contains 39.3% sodium. Calculate the mass of:
c.   sodium chloride that contains 5 g of sodium
d.   sodium that will be combined with 2.75 g of chlorine

could someone show me the step to working this out ???
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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #7936 on: March 16, 2019, 07:24:00 pm »
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Hey

So from my understanding, biodiesel that is derived from triglycerides that are unsaturated tend to have a higher viscosity compared to those that are saturated? Because petrodiesel is nonpolar and usually contains only single bonds between its carbons and its resistance to flow at lower temperatures is relatively low compared to biodiesel which is thicker. So I don't get why unsaturated biodiesel (which only contains single bond triglycerides) has a higher resistance to flow at lower temperatures than saturated biodiesel (which contains double bonds) when petrodiesel (which only contains single bonds) has a lower resistance to flow compared to biodiesel. Shouldn't unsaturated triglyceride based biodiesel have a lower viscosity than saturated due to the weaker intermolecular bonds?



Do you have a source for this? I would believe that the unsaturated would have a lower viscosity and the data that I've looked at seems to be consistent with this.


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Re: VCE Chemistry Question Thread
« Reply #7937 on: March 16, 2019, 08:52:42 pm »
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Do you have a source for this? I would believe that the unsaturated would have a lower viscosity and the data that I've looked at seems to be consistent with this.


Nvm, I think I read the textbook incorrectly. Yeah that sounds right since petrodiesel which contains single bonds also is less viscous than biodiesel which has double bonds
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dream chaser

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Re: VCE Chemistry Question Thread
« Reply #7938 on: March 17, 2019, 05:15:48 pm »
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Hi Guys,

Need help with this question. The question is in the attachment.

What do they mean faradays of charge? Is that just F? And I thought 1F=96500C/mol?

I'm confused. How will you solve them?

Thanks. All help will be really much appreciated.

persistent_insomniac

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Re: VCE Chemistry Question Thread
« Reply #7939 on: March 18, 2019, 06:35:10 pm »
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Just to confirm - are gas laws/relationships between factors (e.g. temp, pressure, mol) affecting gases, application of gases etc on the study design? Or do we just have to know how to use the gas formulus (PV = nRT?)

Owlbird83

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Re: VCE Chemistry Question Thread
« Reply #7940 on: March 18, 2019, 08:59:43 pm »
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If a question on galvanic cells asks for the electrode/cathode/anode on a half cell where the electrode is aqueous, is it better to write the aqueous electrode or to just say an inert electrode?
Thanks :)
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Agimo

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Re: VCE Chemistry Question Thread
« Reply #7941 on: March 19, 2019, 06:26:27 pm »
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Hey guys,

What are some reasons for why the experimental voltage of a galvanic cell is lower then the expected voltage shown on the electrochemical series?

TylerD9

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Re: VCE Chemistry Question Thread
« Reply #7942 on: March 19, 2019, 10:07:45 pm »
+1
Hey guys,

What are some reasons for why the experimental voltage of a galvanic cell is lower then the expected voltage shown on the electrochemical series?

It will be different if it isn't under standard conditions, i.e. 25 degrees, 1M, 100kPa etc
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #7943 on: March 19, 2019, 10:11:00 pm »
+3
Hi Guys,

Need help with this question. The question is in the attachment.

What do they mean faradays of charge? Is that just F? And I thought 1F=96500C/mol?

I'm confused. How will you solve them?

Thanks. All help will be really much appreciated.
The question is a little misleading. For this question, just pretend 1 F = 96500 C, even though it's not.

Just to confirm - are gas laws/relationships between factors (e.g. temp, pressure, mol) affecting gases, application of gases etc on the study design? Or do we just have to know how to use the gas formulus (PV = nRT?)
You should be able to work out the relationships from the ideal gas law, which was formulated to combine all of the factors together in one equation.

If a question on galvanic cells asks for the electrode/cathode/anode on a half cell where the electrode is aqueous, is it better to write the aqueous electrode or to just say an inert electrode?
Thanks :)
Electrodes are never aqueous though......they're always solid. They might be inert, like platinum, but they're never aqueous. Otherwise, you can't separate any products from the reactants if the reactions occur in solution.

Hey guys,

What are some reasons for why the experimental voltage of a galvanic cell is lower then the expected voltage shown on the electrochemical series?
Non-standard conditions, faulty voltmeter, extra resistance in the system, for starters.
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JR_StudyEd

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Re: VCE Chemistry Question Thread
« Reply #7944 on: March 20, 2019, 05:28:07 pm »
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Hi, why is it that in an exothermic reaction, the products have stronger bonds than the reactants?
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Owlbird83

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Re: VCE Chemistry Question Thread
« Reply #7945 on: March 20, 2019, 06:44:10 pm »
+1
Hi, why is it that in an exothermic reaction, the products have stronger bonds than the reactants?

Are you sure about this, because I was taught the opposite. In an exothermic reaction, the reactants have higher energy bonds than the products because some energy is released as heat in the reaction
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JR_StudyEd

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Re: VCE Chemistry Question Thread
« Reply #7946 on: March 20, 2019, 07:07:51 pm »
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Are you sure about this, because I was taught the opposite. In an exothermic reaction, the reactants have higher energy bonds than the products because some energy is released as heat in the reaction
I'll write out the question in full:

(From Checkpoints 2019)
Questions 2
The reaction between hydrogen and oxygen is described by the equation

2H2(g) + O2(g) ----> 2H2O(g); Delta H = -484kJ mol-1

(b) is where my question is. It says: "describe how the strength of the bonds in the reactants is related to the strength of the bonds in the products". How would you answer this?

Because I had no idea how to answer the question, I first checked the answers. It said "In an exothermic reaction, the products are at lower energy and have stronger bonds than the reactants.". Why is that so?
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turtlebanana

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Re: VCE Chemistry Question Thread
« Reply #7947 on: March 23, 2019, 02:14:01 pm »
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Hey guys,

- When talking about equilibrium, is it true that both Qc and Kc involve concentration of products/concentration of reactants, but Qc is the calculation when the mixture is not at equilibrium, whereas Kc is the calculation when the mixture is always at equilibrium?

- What does the subscript c next to Q and K represent? When do we include it and can there be different subscripts?

- What does it mean by '' for a particular reaction, Kc is constant for all equilibrium mixtures at a fixed temperature''? (what particular reaction are they talking about? isn't Kc always constant at equilibrium and is only affected by the temperature?)

Thanks
« Last Edit: March 23, 2019, 02:18:51 pm by turtlebanana »
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leiva-acuna.s

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Re: VCE Chemistry Question Thread
« Reply #7948 on: March 24, 2019, 01:50:06 pm »
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Hi,

Could anyone please unpack what this dot point of the study design requires us to know? I can't find anything that explicitly refers to this dot point. Does this dot point include Equilibrium Law?

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rani_b

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Re: VCE Chemistry Question Thread
« Reply #7949 on: March 24, 2019, 07:31:30 pm »
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b) is where my question is. It says: "describe how the strength of the bonds in the reactants is related to the strength of the bonds in the products". How would you answer this?

Because I had no idea how to answer the question, I first checked the answers. It said "In an exothermic reaction, the products are at lower energy and have stronger bonds than the reactants.". Why is that so?

I think because:
In an exothermic reaction, it takes less energy (the activation energy is less) to break the bonds in the reactants than is released when new bonds form in the products.
In an endothermic reaction, it's the opposite way: so it takes more energy to break bonds in the reactants than is released when new bonds form in the products.
Someone just confirm if I'm right because I don't wanna be saying wrong information...   :o
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