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October 21, 2025, 06:24:29 pm
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Author Topic: TrueTears question thread  (Read 33722 times)  Share 

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jackinthepatch

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Re: TrueTears question thread
« Reply #15 on: January 17, 2009, 06:51:10 pm »
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I'm sure somebody could sum that up much better than I did lol, so feel free to.
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TrueTears

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Re: TrueTears question thread
« Reply #16 on: January 17, 2009, 06:57:40 pm »
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thanks jack XD

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TrueTears

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Re: TrueTears question thread
« Reply #17 on: January 17, 2009, 06:58:03 pm »
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also.

When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of phosphorus is produced. The molar mass of the oxide is

a) determine the empirical formula of the phosphorus oxide.

I did this question and got the answer right, but I'm not quite sure if some of my working is correct, could someone please check.

My working is as follows:

















(which is rounded to 10)

so the formula of the oxide is , but since it askes for empirical, we must cancel down and we get . Which is the correct answer.

But i'm not sure about my first line of working, why does the reactant Phosphorus have the same subscript as the one in Oxygen? When they combine couldn't the subscript change?
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Wizard

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Re: TrueTears question thread
« Reply #18 on: January 17, 2009, 07:49:54 pm »
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If 0.1g is the Phosphorus part, then 0.128g is the Oxygen part.

n(P) = .003225
n(O) = .008

mole ratio:     P:O = 1 : 2.5
                         = 2 : 5

Therefore, empirical formula is P2O5

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TrueTears

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Re: TrueTears question thread
« Reply #19 on: January 17, 2009, 08:18:32 pm »
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Quote from: Wizard on January 17, 2009, 07:49:54 pm
If 0.1g is the Phosphorus part, then 0.128g is the Oxygen part.

n(P) = .003225
n(O) = .008

mole ratio:     P:O = 1 : 2.5
                         = 2 : 5

Therefore, empirical formula is P2O5



ahh much easier way haha thanks wizard
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TrueTears

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Re: TrueTears question thread
« Reply #20 on: January 17, 2009, 10:38:54 pm »
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and another one

What volume of 0.100M sulfuric acid would be required to neutralize in a solution containing 0.500g of sodium hydroxide and 0.800 g of potassium hydroxide.
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vce08

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Re: TrueTears question thread
« Reply #21 on: January 17, 2009, 10:45:21 pm »
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find out the mol of OH- ions in the solution (added up from the number of mols of NaOH and KOH)

Halve this number and you will have the number of mol of H2SO4 required to neutralize.
then use n = cV to find out the required volume of H2SO4
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TrueTears

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Re: TrueTears question thread
« Reply #22 on: January 17, 2009, 10:58:36 pm »
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Quote from: Table on January 17, 2009, 10:45:21 pm
find out the mol of OH- ions in the solution (added up from the number of mols of NaOH and KOH)

Halve this number and you will have the number of mol of H2SO4 required to neutralize.
then use n = cV to find out the required volume of H2SO4


thanks i just got it anyway haha
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TrueTears

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Re: TrueTears question thread
« Reply #23 on: January 17, 2009, 10:59:03 pm »
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Also

Write the ionic equation to



How i work out ionic equation is break everything up into their ions and 'cancel' out the identical ions on the LHS and RHS of the equation.

so the equation becomes:


the and 'cancels' and we are left with

However my book's answers say its:
How do you get that? And why didn't the 'cancel' ?

Many thanks !
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vce08

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Re: TrueTears question thread
« Reply #24 on: January 17, 2009, 11:01:20 pm »
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Because the potassium ion is simply a spectator ion which does not participate in the reaction.
Not sure about the ethanoate ion though
« Last Edit: January 17, 2009, 11:03:25 pm by Table »
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TrueTears

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Re: TrueTears question thread
« Reply #25 on: January 17, 2009, 11:10:38 pm »
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yeah i know they cancelled out the , but why not the ion?
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TrueTears

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Re: TrueTears question thread
« Reply #26 on: January 18, 2009, 02:24:45 am »
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just another 2 Q's.

1. Concentrated solutions of NaOH are not used in a titration, explain why.

2. Standard sodium hydroxide solutions can not be prepared directly from the solid. Explain why.
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vce08

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Re: TrueTears question thread
« Reply #27 on: January 18, 2009, 02:29:22 am »
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1. If the concentrated NaOH is used, if the titration is out by a single drop, this will have a profound effect on the calculations of number of mol NaOH titrated than if a lower concentration of NaOH is used.
2. NaOH reacts with the acidic carbon dioxide in the air so we cannot know the number of mol of NaOH for a given weight of the solid.
« Last Edit: January 18, 2009, 02:34:03 am by Table »
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TrueTears

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Re: TrueTears question thread
« Reply #28 on: January 18, 2009, 02:53:44 am »
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ahh thanks table
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TrueTears

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Re: TrueTears question thread
« Reply #29 on: January 18, 2009, 02:56:52 am »
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yeah and also for that other Q

Quote from: TrueTears on January 17, 2009, 10:59:03 pm
Also

Write the ionic equation to



How i work out ionic equation is break everything up into their ions and 'cancel' out the identical ions on the LHS and RHS of the equation.

so the equation becomes:


the and 'cancels' and we are left with

However my book's answers say its:
How do you get that? And why didn't the 'cancel' ?

Many thanks !


anyone know why the don't cancel?
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