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June 25, 2021, 01:35:54 am

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#### wingdings2791

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##### Re: VCE Chemistry Question Thread
« Reply #9105 on: May 05, 2021, 11:07:57 pm »
+5
Hi Chocolatepistachio,
I suggest using a variable to represent the unknown concentrations. As the experiment is investigating the decomposition of NOCl(g) and the initial amount of NOCl(g) is given, the initial:
$n(NOCl) = 1.00 mol$
$n(NO) = 0 mol$
$n(Cl_{2}) = 0 mol$
As it is unknown how much NOCl(g) is reacted to form NO(g) and Cl2(g), we can use a variable to represent the change in mole. This change will be proportional to the NOCl:NO:Cl2 ratio of 2:2:1.
$Δn(NOCl) = -2x$
$Δn(NO) = +2x$
$Δn(Cl_{2}) = +x$
meaning that since the volume of the flask is 2.00L, equilibrium moles will be
$n(NOCl)=\frac{1.00 mol-2x}{2.00L}$
$n(NO) = \frac{2x}{2.00L}$
$n(Cl_{2}) = \frac{x}{2.00L}$
Now the equilibrium constant. Following
$K=\frac{[products]^{coefficients}}{[reactants]^{coefficients}}$
and the given equilibrium constant (at 35oC, presumably the temperature of this experimental setup)
we can construct the equation $K=\frac{[NO]^2[Cl]}{[NOCl]^2}=\frac{x^2(0.5x)}{(0.5-x)^2}$
from which we can get $\frac{0.5x^3}{(0.5-x)^2}=1.60\times10^{-5}$
$\frac{x^3}{(0.5-x)^2}=2.20\times10^{-5}$
Applying polynomials, you would eventually come to $x=\frac{2.2}{16}\times10^{-1}=1.375\times10^{-2}$
which can be substituted back into the original equilibrium mole values to get our final answers of:
$n(NOCl)=\frac{1.00 mol-2x}{2.00L}=0.48625M=4.9\times10^{-1}M$
$n(NO) = \frac{2x}{2.00L}=0.01375M=1.4\times10^{-2}M$
$n(Cl_{2}) = \frac{x}{2.00L}=0.006875M=6.9\times10^{-3}M$
at equilibrium.

Hope this helps, please let me know if there are any errors
2019- Chinese SL [42]
2020- Biology [43] Music Performance [49]
2021- Chemistry, Methods, English Language

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #9106 on: May 07, 2021, 02:32:32 pm »
0
Isn’t the equilibrium concentration for NOCl 0.500-2x , And NO 2x

#### fun_jirachi

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##### Re: VCE Chemistry Question Thread
« Reply #9107 on: May 07, 2021, 06:59:35 pm »
+3
Isn’t the equilibrium concentration for NOCl 0.500-2x , And NO 2x

No, since the entire numerator has to be divided by the denominator. $\frac{1-2x}{2} = 0.5-x$, not $0.5-2x$. This also applies for the concentration of NO.
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#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #9108 on: May 07, 2021, 07:06:21 pm »
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but itsn't the equilibrium concentration just the initial + change like this

#### Corey King

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##### Re: VCE Chemistry Question Thread
« Reply #9109 on: May 07, 2021, 07:58:40 pm »
0
Hey guys,

https://snipboard.io/26fX9O.jpg

#### wingdings2791

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##### Re: VCE Chemistry Question Thread
« Reply #9110 on: May 07, 2021, 10:47:11 pm »
+3
Hey guys,

https://snipboard.io/26fX9O.jpg

Hi Corey King,
Electrodes can be positive or negative depending on each electrode's proportion of electrons. It's the same case with electrolytic cells; whilst galvanic cells have a positive cathode (electrodes consumed in reduction, therefore net decrease in free electrons) and negative anode (electrodes produced in oxidation, therefore net increase in free electrons), electrolytic cells have a negative cathode (electrons pushed towards the cathode by power supply) and a positive anode (electrons produced by oxidation are transferred to the cathode by the power supply, leaving the positive ions from oxidation at the anode).

As you've described, the oxidation of Cu(s) will release Cu2+(aq) and electrons into the solution at the anode, which on its own does not have an effect on overall charge. The continuous transport of electrons from the anode to the cathode by the power supply maintains the positive charge of the anode. If you like, you can think about it as the electrons produced by oxidation being transferred to the cathode so they can be consumed in reduction.
2019- Chinese SL [42]
2020- Biology [43] Music Performance [49]
2021- Chemistry, Methods, English Language

#### Corey King

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##### Re: VCE Chemistry Question Thread
« Reply #9111 on: May 08, 2021, 07:28:23 am »
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So why is the charge positive from the anode if it's producing e-? Which are negative

#### ArtyDreams

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##### Re: VCE Chemistry Question Thread
« Reply #9112 on: May 08, 2021, 09:16:26 am »
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So why is the charge positive from the anode if it's producing e-? Which are negative

If you take a standard electric circuit, the current moves out of the positive side into the negative side. This is the same case here, where the electrons need to flow out of the positive side into the negative anode. This is why the anode is positive in an electrolytic cell, as the current always flows from anode to cathode.

#### wingdings2791

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##### Re: VCE Chemistry Question Thread
« Reply #9113 on: May 08, 2021, 02:06:41 pm »
+3
but itsn't the equilibrium concentration just the initial + change like this
What you've labeled as equilibrium concentration is actually equilibrium mole ($n$ of each reactant/product), not concentration. You need to divide by the volume (which is given as 2.00L in the question) to obtain equilibrium concentrations.
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#### Corey King

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##### Re: VCE Chemistry Question Thread
« Reply #9114 on: May 08, 2021, 03:52:26 pm »
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If you take a standard electric circuit, the current moves out of the positive side into the negative side. This is the same case here, where the electrons need to flow out of the positive side into the negative anode. This is why the anode is positive in an electrolytic cell, as the current always flows from anode to cathode.

Ohh I see, so positive refers to Direction of electron flow, not the Charge of the current that is flowing?

The the anode has negative charge (is this polarity) and positive charge in its electrode for a primary cell,

But in a secondary cell the anode is switched to the opposite electrode, and the electrode it now exists with is still is still providing electron Current so the current is positive?

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #9115 on: May 11, 2021, 01:52:34 pm »
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For this question at point A would the volume have decreased (increasing pressure) and reactions shifts to the left

At point B would the temperature have decreased, and the reaction shifts to the left

At point c would some C2H6 have been added and the reactions shifts to the left

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #9116 on: May 11, 2021, 04:57:45 pm »
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for this question
a) if the temperature is decreased would it shift to the left
if the volume of the reaction vessel is decreased, increasing pressure would it shift to the left
c) if an inert gas(neon) is added at constant volume would it shift to the right where there are more moles of gas

#### beep boop

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##### Re: VCE Chemistry Question Thread
« Reply #9117 on: May 13, 2021, 12:45:51 pm »
0
does anyone have any tips on making a scientific poster. my research investigation task is coming soon and im a bit worried for it.  ive done plenty of scientific posters, however, i always get a mediocre grade.

any help is appreciated!

beep boop over and out
class of '22

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #9118 on: May 13, 2021, 03:40:44 pm »
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Why does only temperature affect kc

#### Chocolatepistachio

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##### Re: VCE Chemistry Question Thread
« Reply #9119 on: May 13, 2021, 04:09:43 pm »
0
Hi Chocolatepistachio,
I suggest using a variable to represent the unknown concentrations. As the experiment is investigating the decomposition of NOCl(g) and the initial amount of NOCl(g) is given, the initial:
$n(NOCl) = 1.00 mol$
$n(NO) = 0 mol$
$n(Cl_{2}) = 0 mol$
As it is unknown how much NOCl(g) is reacted to form NO(g) and Cl2(g), we can use a variable to represent the change in mole. This change will be proportional to the NOCl:NO:Cl2 ratio of 2:2:1.
$Δn(NOCl) = -2x$
$Δn(NO) = +2x$
$Δn(Cl_{2}) = +x$
meaning that since the volume of the flask is 2.00L, equilibrium moles will be
$n(NOCl)=\frac{1.00 mol-2x}{2.00L}$
$n(NO) = \frac{2x}{2.00L}$
$n(Cl_{2}) = \frac{x}{2.00L}$
Now the equilibrium constant. Following
$K=\frac{[products]^{coefficients}}{[reactants]^{coefficients}}$
and the given equilibrium constant (at 35oC, presumably the temperature of this experimental setup)
we can construct the equation $K=\frac{[NO]^2[Cl]}{[NOCl]^2}=\frac{x^2(0.5x)}{(0.5-x)^2}$
from which we can get $\frac{0.5x^3}{(0.5-x)^2}=1.60\times10^{-5}$
$\frac{x^3}{(0.5-x)^2}=2.20\times10^{-5}$
Applying polynomials, you would eventually come to $x=\frac{2.2}{16}\times10^{-1}=1.375\times10^{-2}$
which can be substituted back into the original equilibrium mole values to get our final answers of:
$n(NOCl)=\frac{1.00 mol-2x}{2.00L}=0.48625M=4.9\times10^{-1}M$
$n(NO) = \frac{2x}{2.00L}=0.01375M=1.4\times10^{-2}M$
$n(Cl_{2}) = \frac{x}{2.00L}=0.006875M=6.9\times10^{-3}M$
at equilibrium.

Hope this helps, please let me know if there are any errors

hey sorry how did you work out x using polynomials from the x3/(0.5-x)2= 2.20 x 10-5