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March 28, 2024, 07:32:29 pm

Author Topic: VCE Chemistry Question Thread  (Read 2312962 times)  Share 

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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #30 on: January 04, 2014, 10:42:40 am »
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When a certain non-metal whose formula is X8 burns in air XO3 forms. Write a balanced equation for this reaction. If 120.0g of oxygen gas is consumed completely, along with 80.0g of X8, identify element X.

Is this balanced equation correct: X8 (g) + 12 O2 (g) -> 8XO3 (g)
I'm not sure how to proceed with this question....

Your help is appreciated   :D

Sorry. The chemical formula makes it look slightly obvious as to what it is :P

Just saying, X8 looks like a solid to me. But you're not meant to know that.

Your equation is balanced correctly though. Now, you know that 120.0 g of oxygen gas is used up. This is 120/32=15/4 moles of oxygen gas (divide by 32 not 16 remember). This must have reacted with exactly 15/48 moles of X8 by mole ratios. If this mass of X8 is 80 grams, one mole of your unknown compound is 80*48/15=80*16/5=256 grams
This is for X8, so the molar mass of X is 32 grams
AKA sulfur


How many non-metals are there anyway that exist as X8 and form trioxides? :P
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brightsky

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Re: VCE Chemistry Question Thread
« Reply #31 on: January 04, 2014, 10:46:36 am »
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Sulfur fits the bill quite nicely. Let see if it is indeed sulfur...

n(O2) = m/M = 120.0/(2*16.0) = 120.0/32.0 = 3.75 mol
n(X8) = m/M = 80.0/M(X8) = 3.75/12 = 0.3125 mol (since the n(X8) : n(O2) = 1 : 12)
M(X8) = 80.0/0.3125 = 256 g/mol
M(X) = 256/8 = 32 g/mol

The element X is indeed sulfur.
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #32 on: January 04, 2014, 10:55:23 am »
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X8 reacts completely with oxygen; a combustion reaction.

X8(s) + 12O2(g) ---> 8XO3(s)

Now that we have our balanced equation, we use stoichiometry.

We know that we've reacted 120.0g of O2 with 80g of X8.

n(O2): n(X8)
12: 1
120/32: n
Cross multiply
n(X8) = 0.3125 mol

We want to find the molecular mass of this metal, so.

0.3125 = 80 / mr
Mr(X8) = 256
Mr(X) = 256 / 8
Mr(X) = 32

Thus, by consulting our periodic table, we realise the metal is Sulphur.


DJA

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Re: VCE Chemistry Question Thread
« Reply #33 on: January 04, 2014, 11:11:52 am »
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Sorry. The chemical formula makes it look slightly obvious as to what it is :P

AKA sulfur

How many non-metals are there anyway that exist as X8 and form trioxides? :P

lol

Nliu--> Looks at the question...A few seconds later knows what the answer is.
Who bothers to calculate these days anyway?
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eagles

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Re: VCE Chemistry Question Thread
« Reply #34 on: January 04, 2014, 11:22:49 am »
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Haha thanks for all your help and different ways to approaching the question.

I have another question:

A 10.0 mL sample of HNO3 was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M KOH solution. What was the concentration of the original nitric acid?



DJA

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Re: VCE Chemistry Question Thread
« Reply #35 on: January 04, 2014, 12:30:12 pm »
+2
Haha thanks for all your help and different ways to approaching the question.
I have another question:
A 10.0 mL sample of HNO3 was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M KOH solution. What was the concentration of the original nitric acid?

The equation for the neutralisation reaction from the info is this:
HNO3 + KOH ---> KNO3 +H2O

The mole ratio between HNO3 and KOH is 1:1 meaning that basically 1 mol of HNO3 will react with 1 mol of KOH. Keep this ratio in mind.

Transform everything to moles for calculation:
n(KOH) in moles = 0.60 * 0.050 = 0.030mol KOH reacted

Remember the mole ratio? We now know that:
n(HNO3) = n(KOH) = 0.030mol  due to the 1:1 ratio

Now this 0.030mol is from 0.025L of solution. Hence in 0.100 L of solution we use a multiplication factor of 4 to get the amount in mol in the 100mL.
n(HNO3) in 100.0mL = 4 * 0.030mol = 0.120 mol

This amount is the total mol of HNO3 from the very start (which was originally in only 10mL of solution). Therefore we now can easily calculate the concentration of the original HNO3.
C(HNO3) original = 0.120 / 0.010 = 12M concentration for the nitric acid.

(sheesh that's some strong nitric acid...now I am less sure my answer is right)
« Last Edit: January 04, 2014, 12:32:43 pm by DJALogical »
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eagles

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Re: VCE Chemistry Question Thread
« Reply #36 on: January 04, 2014, 01:04:24 pm »
+1
Yep that's the correct answer thank you!


(sheesh that's some strong nitric acid...now I am less sure my answer is right)

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Re: VCE Chemistry Question Thread
« Reply #37 on: January 04, 2014, 01:09:16 pm »
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Yep that's the correct answer thank you!

You're very welcome.
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eagles

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Re: VCE Chemistry Question Thread
« Reply #38 on: January 04, 2014, 01:55:52 pm »
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How would you draw compound C3H3O6?


lzxnl

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Re: VCE Chemistry Question Thread
« Reply #39 on: January 04, 2014, 02:05:59 pm »
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How would you draw compound C3H3O6?

That's not a stable molecule; odd number of electrons and organic molecules like that aren't stable
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Re: VCE Chemistry Question Thread
« Reply #40 on: January 04, 2014, 06:01:56 pm »
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Thanks! But it's the answer I have for this question:

Compound Z consists of 26.7% carbon, 2.2% hydrogen and 71.1% oxygen.
Its molecular mass is 90.0g mol-1.

Can someone please clarify?

Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #41 on: January 04, 2014, 06:19:04 pm »
0
Thanks! But it's the answer I have for this question:

Compound Z consists of 26.7% carbon, 2.2% hydrogen and 71.1% oxygen.
Its molecular mass is 90.0g mol-1.

Can someone please clarify?

That compound's empirical formula is CHO2

The compound's molecular formula is C2H2O4

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Re: VCE Chemistry Question Thread
« Reply #42 on: January 04, 2014, 06:24:25 pm »
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Thanks! It makes more sense now

Professor_Oak

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Re: VCE Chemistry Question Thread
« Reply #43 on: January 05, 2014, 12:24:15 pm »
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Hi guys can somebody explain the logic behind 25c in a relatively detailed manner? I can sort of see how it may work but I'm still a bit confused with this sort of question.
« Last Edit: January 05, 2014, 12:25:57 pm by Professor_Oak »
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Re: VCE Chemistry Question Thread
« Reply #44 on: January 05, 2014, 12:55:05 pm »
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We know in the 20mL of CH3CH2OH we had 0.000690 mol
We want to figure out how many mols of CH3CH2OH was in the initial 250mL used.

The way I did these questions was setting it up like a sort of ratio.

0.000690 mol -> 20mL
 x mol -> 250mL

x =0.000690 * (250/20)
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