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Author Topic: TrueTears question thread  (Read 33907 times)  Share 

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TrueTears

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TrueTears question thread
« on: January 16, 2009, 11:21:09 pm »
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An impure sample of iron(II) sulfate, weighing 1.545 g , was treated to produce a precipitate of . If the mass of the dried precipitate was 0.315 g, calculate the percentage of iron in the sample.
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Mao

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Re: TrueTears question thread
« Reply #1 on: January 16, 2009, 11:50:58 pm »
+1




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TrueTears

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Re: TrueTears question thread
« Reply #2 on: January 17, 2009, 12:02:03 am »
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ah thanks mao

A brand of toothpaste contains 0.22% by mass sodium fluoride (NaF). Calculate the mass of fluoride ions in a tube containing 120g of the paste
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/0

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Re: TrueTears question thread
« Reply #3 on: January 17, 2009, 12:25:05 am »
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To 2 significant figures,
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vce08

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Re: TrueTears question thread
« Reply #4 on: January 17, 2009, 03:20:35 pm »
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Empirical formula would be P2O5.

just like how the empirical formula of the compound C2H6 would be CH3.
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TrueTears

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Re: TrueTears question thread
« Reply #5 on: January 17, 2009, 03:33:39 pm »
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lol i know 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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TrueTears

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Re: TrueTears question thread
« Reply #6 on: January 17, 2009, 03:55:03 pm »
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and also

For a 0.20 M solution of , calculate the amount, in mol, of ions.
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mark_alec

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Re: TrueTears question thread
« Reply #7 on: January 17, 2009, 04:17:57 pm »
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Quote from: TrueTears on January 17, 2009, 03:55:03 pm
For a 0.20 M solution of , calculate the amount, in mol, of ions.

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TrueTears

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Re: TrueTears question thread
« Reply #8 on: January 17, 2009, 04:30:33 pm »
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Quote from: mark_alec on January 17, 2009, 04:17:57 pm
Quote from: TrueTears on January 17, 2009, 03:55:03 pm
For a 0.20 M solution of , calculate the amount, in mol, of ions.



so is the amount in mol 0.4? my book says 0.1 mol.
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Wizard

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Re: TrueTears question thread
« Reply #9 on: January 17, 2009, 05:38:18 pm »
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Did they give you a volume? Because there is probably not enough info if they didnt
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TrueTears

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Re: TrueTears question thread
« Reply #10 on: January 17, 2009, 05:43:26 pm »
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Quote from: Wizard on January 17, 2009, 05:38:18 pm
Did they give you a volume? Because there is probably not enough info if they didnt

nope they didn't give any volume.
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/0

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Re: TrueTears question thread
« Reply #11 on: January 17, 2009, 05:54:03 pm »
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Quote from: TrueTears on January 17, 2009, 05:43:26 pm
Quote from: Wizard on January 17, 2009, 05:38:18 pm
Did they give you a volume? Because there is probably not enough info if they didnt

nope they didn't give any volume.

The book made a mistake, they left out the volume. It is meant to be 250mL
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TrueTears

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Re: TrueTears question thread
« Reply #12 on: January 17, 2009, 05:58:00 pm »
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Quote from: /0 on January 17, 2009, 05:54:03 pm
Quote from: TrueTears on January 17, 2009, 05:43:26 pm
Quote from: Wizard on January 17, 2009, 05:38:18 pm
Did they give you a volume? Because there is probably not enough info if they didnt

nope they didn't give any volume.

The book made a mistake, they left out the volume. It is meant to be 250mL

thanks, also wat about that first line of working query?
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TrueTears

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Re: TrueTears question thread
« Reply #13 on: January 17, 2009, 06:08:39 pm »
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also

A student wishes to prepare 500mL of a standard solution of any base of concentration of 0.2500M
Would it be better to prepare the solution using solid sodium hydroxide or anhydrous sodium carbonate? Explain.
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jackinthepatch

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Re: TrueTears question thread
« Reply #14 on: January 17, 2009, 06:49:48 pm »
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Be better to prepare using anhydrous sodium carbonate. Explanation: anhydrous sodium carbonate is a primary standard, meaning it will not react with the air, making it very pure, which in turn makes its number of mole able to be accurately calculated from it's mass. (It's number of mole would be calculated more accurately and reliably than than sodium hydroxide). This would ensure that the concentration of the 500mL solution is as accurate as possible.
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