[mr cool]

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++]

Dear reezyfbaby;

Do it analog the example in:   http://en.wikipedia.org/wiki/Graham's_law
(Wiki is "always" your friend.)

You did wrong rearrange your equation!

Good Luck!
                    ARGOS⁺⁺

[mr cool]

Dear reezyfbaby;

Do it analog the example in:   http://en.wikipedia.org/wiki/Graham's_law
(Wiki is "always" your friend.)

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++]

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]

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++]

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]

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++]

Dear reezyfbaby;

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

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

Good Luck!
                    ARGOS⁺⁺

[mr cool]

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++]

Dear reezyfbaby;

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

Good Luck!
                    ARGOS⁺⁺

[mr cool]

Dear reezyfbaby;

Sorry!, - I get another value for MW_UK.
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++]

Dear reezyfbaby;

Let’s start to rearrange the original law for MW_UK:
                     ______          ______
v_UK / v_CO2 =  \| MW_CO2  /  \| MW_UK

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

Good Luck!
                    ARGOS⁺⁺

[mr cool]

Dear reezyfbaby;

Let’s start to rearrange the original law for MW_UK:
                     ______          ______
v_UK / v_CO2 =  \| MW_CO2  /  \| MW_UK

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

Good Luck!
                    ARGOS⁺⁺

What is v on the left?

[ARGOS++]

Dear reezyfbaby;

v on the left side are the speeds.

Good Luck!
                    ARGOS⁺⁺

[mr cool]

So....
                      ____      _____
V uk / V co2 = \| uk  /  \|44.0095  g co2
V uk / v co2 = uk g / 44.0085 g co2