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### Topic: Graham's law practice  (Read 10287 times)

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#### mr cool

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##### Graham's law practice
« on: February 16, 2009, 06:00:59 PM »
Ok so i gave this problem a try... an unknown gas is found to travel twice as fast as CO2, which traveled 14.00 cm in 12.00 sec.  What is the molecular mass of the unknown gas?

My work....

(28.00 cm / 16.00 sec).     _____
------------------------  = \| UK.           (square root)
(14.00 cm / 16.00 sec).   ---------
\| 44.0095 g/mol
_______                            _____________
( \| UK      ) (14.00 cm)        (\|44.0095 g/ mol) (28.00 cm)
---------------------------  =     --------------------------------
16.00 sec.                           14.00 cm

____             ____________
\|  UK     =   (\|44.0095 g/mol) (28.00 cm)
------------------------------
14.00 cm

I dont know what to do with my final set-up to get the answer help?

UK = unknown btw

#### ARGOS++

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##### Re: Graham's law practice
« Reply #1 on: February 16, 2009, 06:10:10 PM »
Dear reezyfbaby;

Do it analog the example in:   http://en.wikipedia.org/wiki/Graham's_law

You did wrong rearrange your equation!

Good Luck!
ARGOS++
« Last Edit: February 16, 2009, 06:21:43 PM by ARGOS++ »

#### mr cool

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##### Re: Graham's law practice
« Reply #2 on: February 16, 2009, 07:24:23 PM »
Dear reezyfbaby;

Do it analog the example in:   http://en.wikipedia.org/wiki/Graham's_law

You did wrong rearrange your equation!

Good Luck!
ARGOS++

So which part did i do wrong? I followed the example elwe did in class.

#### ARGOS++

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##### Re: Graham's law practice
« Reply #3 on: February 16, 2009, 07:32:17 PM »

Dear reezyfbaby;

Please check your first equation for which gas has the higher speed?
(And check also if 16 sec or 12 sec will be the correct one.)

Good Luck!
ARGOS++

#### mr cool

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##### Re: Graham's law practice
« Reply #4 on: February 16, 2009, 07:36:07 PM »

Dear reezyfbaby;

Please check your first equation for which gas has the higher speed?
(And check also if 16 sec or 12 sec will be the correct one.)

Good Luck!
ARGOS++

Oh poo, it would help if i read the problem more closely.  The UK is lighter. But why is it 12 because dont you take the total ammount of time it took?

#### ARGOS++

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##### Re: Graham's law practice
« Reply #5 on: February 16, 2009, 07:41:32 PM »

Dear reezyfbaby;

The total time of 12 sec. would be ok, but it seems that you changed it from your question till your first equation into 16 sec, but why?

Good Luck!
ARGOS++

#### mr cool

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##### Re: Graham's law practice
« Reply #6 on: February 16, 2009, 07:44:17 PM »

Dear reezyfbaby;

The total time of 12 sec. would be ok, but it seems that you changed it from your question till your first equation into 16 sec, but why?

Good Luck!
ARGOS++

I think i meant to make it 18 not 16 but because 12.00 sec + 6.00 sec (time it took for UK because it was 2x faster, or can you not do that?

#### ARGOS++

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##### Re: Graham's law practice
« Reply #7 on: February 16, 2009, 07:49:10 PM »
Dear reezyfbaby;

No!, - that’s definitely not allowed!  Why not using the normal speed?

Additionally Hint: Why not first rearrange the law for MWUK?

Good Luck!
ARGOS++
« Last Edit: February 16, 2009, 08:00:11 PM by ARGOS++ »

#### mr cool

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##### Re: Graham's law practice
« Reply #8 on: February 16, 2009, 08:01:21 PM »
So i did it again and wound up with this as my final equation....
_____________
(\|44.0095 g/ mol)(14.00 cm)
---------------------------------
12.00 sec

I got 7.7 g when i did it all out.

#### ARGOS++

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##### Re: Graham's law practice
« Reply #9 on: February 16, 2009, 08:06:43 PM »

Dear reezyfbaby;

Sorry!, - I get another value for MWUK.
Please check at least your result with the formula of the example in the link I gave you.

Good Luck!
ARGOS++

#### mr cool

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##### Re: Graham's law practice
« Reply #10 on: February 16, 2009, 08:09:20 PM »

Dear reezyfbaby;

Sorry!, - I get another value for MWUK.
Please check at least your result with the formula of the example in the link I gave you.

Good Luck!
ARGOS++

I qoild but i didnt understand what the 2/1 and 2/2 were

#### ARGOS++

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##### Re: Graham's law practice
« Reply #11 on: February 16, 2009, 08:22:46 PM »

Dear reezyfbaby;

Let’s start to rearrange the original law for MWUK:
______          ______
vUK / vCO2 =  \| MWCO2  /  \| MWUK

Are you able to square the law to remove all square roots?

Good Luck!
ARGOS++

#### mr cool

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##### Re: Graham's law practice
« Reply #12 on: February 16, 2009, 08:26:13 PM »

Dear reezyfbaby;

Let’s start to rearrange the original law for MWUK:
______          ______
vUK / vCO2 =  \| MWCO2  /  \| MWUK

Are you able to square the law to remove all square roots?

Good Luck!
ARGOS++

What is v on the left?

#### ARGOS++

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##### Re: Graham's law practice
« Reply #13 on: February 16, 2009, 08:29:10 PM »

Dear reezyfbaby;

v on the left side are the speeds.

Good Luck!
ARGOS++

#### mr cool

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##### Re: Graham's law practice
« Reply #14 on: February 16, 2009, 08:35:46 PM »
So....
____      _____
V uk / V co2 = \| uk  /  \|44.0095  g co2
V uk / v co2 = uk g / 44.0085 g co2