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Offline miriam0920

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I need help with a chemistry problem
« on: 12/09/2008 21:48:59 »
I can't figure this out:

If total volume of seawater is 1.5 x 10(power of 21)L, and  Mg found in seawater is 1.3g of Mg for every kilogram of seawater, calcute how many Liters of seawater is need to extract 8.0 x 10 (power of 4) tons of Mg?

Please help me solve this problem by going step by step.

Thanks.


 

Offline Evie

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« Reply #1 on: 12/09/2008 22:05:40 »
Miriam...

You're not asking us to do your homework for you, are you?  ;)
 

Offline miriam0920

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« Reply #2 on: 12/09/2008 22:13:36 »
No absolutely not.  I tried figuring this one, but I can't seem to add it up.  I have the answer which is 5.5 x 10 (power of 10) L.


Thanks.
 

Offline miriam0920

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« Reply #3 on: 12/09/2008 22:27:27 »
I have the answer yet I want to learn how to calculate something like this. 

I'm burn out.
 

Offline miriam0920

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« Reply #4 on: 12/09/2008 22:28:14 »
Where are my friends Bored Chemist and Lightarrow?  Please help.
 

Offline RD

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« Reply #5 on: 12/09/2008 22:32:52 »
Is "ton"

metric ton = tonne = 1000 kg

imperial ton = 1016 kg

short ton (USA) = 907 kg
« Last Edit: 12/09/2008 23:20:27 by RD »
 

Offline miriam0920

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« Reply #6 on: 12/09/2008 22:35:15 »
RD,  in 1 ton = 2000 lbs so I guess I have to cancel lbs to gram which is 453.6 g/1 lbs.
 

Offline RD

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« Reply #7 on: 12/09/2008 22:51:31 »
A kilogramme of pure water occupies a volume of one litre.

A kilogramme of seawater is slightly less than 1 litre, (0.975 litre), because seawater is more dense than pure water.

Answer is 5.44 x 1010 litres if "ton" = short ton = 907 kg = 2000lbs.
« Last Edit: 12/09/2008 23:04:47 by RD »
 

Offline miriam0920

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« Reply #8 on: 12/09/2008 23:07:50 »
Thank you.  I think I can figure out the steps to solve the problem with the conversion factors that you gave me.

is that I got stuck in that problem and I have an exam on Wednesday.


 

Offline RD

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« Reply #9 on: 12/09/2008 23:14:11 »
Initially I didn't factor-in the increased density of seawater.

The density of seawater varies 1020-1035Kgm3, I used 1025 Kgm3, your question may use a different value.
« Last Edit: 12/09/2008 23:19:14 by RD »
 

Offline miriam0920

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« Reply #10 on: 13/09/2008 00:52:24 »
So that means that if I use your factor I might get it wrong?

Ok so let me go to the original question:

How much L in seawater will I need to extract 8.0 x 10(exponent 4) tons of Magnesium?
The total volume of seawater is 1.5 x 10 (exponent 21) L.
 

Offline RD

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« Reply #11 on: 13/09/2008 01:32:48 »
So that means that if I use your factor I might get it wrong?

1025 Kgm3 appears to be the most commonly used value.
Your question or text book could have a figure anywhere between 1020 and 1035 Kgm3.


To obtain 8.0 x104 (short) tons of Mg,  i.e.  8.0 x104 x 907 Kg = 7.256 x 107 Kg = 7.256 x1010 grams of Mg required.

Your question states there are 1.3 grams of Mg per kilo of seawater, so 7.256 x1010 / 1.3 = 5.58 x1010 Kg of seawater are required.

1 kg of seawater occupies a volume of 0.975 litres, (using density of seawater 1025 Kg m3)

5.58 x1010 Kg of seawater occupies a volume of 5.44 x1010 litres.
« Last Edit: 13/09/2008 01:55:20 by RD »
 

Offline lightarrow

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« Reply #12 on: 13/09/2008 18:41:43 »
I can't figure this out:

If total volume of seawater is 1.5 x 10(power of 21)L, and  Mg found in seawater is 1.3g of Mg for every kilogram of seawater, calcute how many Liters of seawater is need to extract 8.0 x 10 (power of 4) tons of Mg?

Please help me solve this problem by going step by step.

Thanks.

If ton = 1000 kg:

8*104 tons = 8*107 kg = 8*1010 g

8*1010 g/1.3 g = 6.15*1010

this means that you have to take 6.15*1010 kg of seawater

Since V = m/ρ where V = volume, m = mass, ρ = density, taking 1.025 kg/L as ρ(seawater), you have:

V = 6.15*1010 kg/1.025 kg/L = 6.00*1010 L of seawater.

If ton = 907 kg and the book gives 5.5*1010 L as answer, it means he used 1.015 kg/L as ρ(seawater).

Clearly the total volum of seawater is a deceiving information.
« Last Edit: 13/09/2008 18:53:50 by lightarrow »
 

Offline miriam0920

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« Reply #13 on: 14/09/2008 00:44:56 »
Lightarrow, thank you for your answer. I finally wrote the professor so he can tell me if I need to add or what something. 
He gave us a similar problem, however he explained it in the classroom.  I have my first exam Wednesday and I would like to pitch for an A.
However, if the examen is not a "choose the best answer" I will probably end up with the wrong calculation. :(
 

Offline RD

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« Reply #14 on: 14/09/2008 00:58:32 »
To answer the question it is necessary to have a figure for the density of seawater.
This information should be given in question paper, or the textbook.

(As lightarrow said the "total volume of seawater" figure is irrelevant and is included to mislead)
« Last Edit: 14/09/2008 01:05:29 by RD »
 

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« Reply #15 on: 14/09/2008 10:08:47 »
The data given in the question is only specified to 2 significant figures. To that precision, any ton(ne) is 1000Kg, and the density of sea water is 1000Kg/M^3.
 

Offline lightarrow

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« Reply #16 on: 14/09/2008 12:36:56 »
The data given in the question is only specified to 2 significant figures. To that precision, any ton(ne) is 1000Kg, and the density of sea water is 1000Kg/M^3.

Ok for seawater density, but not for ton = 907 kg.
 

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« Reply #17 on: 14/09/2008 13:47:48 »
Sorry, I forgot about the Americans using tons that don't weigh much.
 

Offline miriam0920

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« Reply #18 on: 14/09/2008 18:04:41 »
For us Americans the equivalent to 1 ton = 2000 lbs.  1 lbs (pound) = 453.6g. 
How can I calculate how much seawater is needed?  if 1.3g of Magnesium is in a 1 kg of seawater?
 

Offline RD

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« Reply #19 on: 14/09/2008 18:51:59 »
For us Americans the equivalent to 1 ton = 2000 lbs.  1 lbs (pound) = 453.6g. 
How can I calculate how much seawater is needed?  if 1.3g of Magnesium is in a 1 kg of seawater?


2000lbs = short ton = 907 Kg = 9.07 x 105 grams.
8.0 x104 (short) tons is equal to  (8.0x104)  x (9.07 x 105) grams = 7.256 x1010 grams of Mg required ...
http://www.thenakedscientists.com/forum/index.php?topic=17043.msg194999#msg194999

BTW
Wiki sources give 1.28 grams of Mg per Kg of seawater, rather than 1.3,
this would change my answer (above) of 5.44 x 1010 litres to 5.53 x 1010 litres.
i.e. 5.5 x 1010 rounded off to one decimal place, (2 sig figs), which I believe is the answer you were looking for.
« Last Edit: 14/09/2008 18:59:09 by RD »
 

Offline miriam0920

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« Reply #20 on: 14/09/2008 20:23:52 »
So the 7.256 x 10 to10 g would be divided by 1.3g of Mg and it will give me the 5.5 x 10 to the 10th g needed of seawater?
 

Offline RD

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« Reply #21 on: 14/09/2008 21:22:18 »
So the 7.256 x 10 to10 g would be divided by 1.3g of Mg and it will give me the 5.5 x 10 to the 10th g needed of seawater?
No, that will give you the amount of kilos of seawater required, your question asks for the answer in litres of seawater.

Your question states there are 1.3 grams of Mg per kilo of seawater, so 7.256 x1010 / 1.3 = 5.58 x1010 Kg of seawater are required.

1 kg of seawater occupies a volume of 0.975 litres, (using density of seawater 1025 Kg m3)

So 5.58 x1010 Kg of seawater occupies a volume of 5.44 x1010 litres.

[The answer is 5.5 x1010 litres if you use 1.28g of Mg per Kg instead of 1.3g].
« Last Edit: 14/09/2008 21:27:19 by RD »
 

Offline RD

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« Reply #22 on: 14/09/2008 21:50:19 »
Here are the Wikipedia sources I have used...

Quote
The average density of seawater at the surface of the ocean is 1.025 g/ml [1025 Kgm3] ; seawater is denser than fresh water 
http://en.wikipedia.org/wiki/Seawater

Concentration of Magnesium (Mg) in seawater 0.0528 mole/Kg
http://en.wikipedia.org/wiki/Seawater#Geochemical_explanations

1 mole of Magnesium is 24.3050  grams
http://en.wikipedia.org/wiki/Magnesium

So there are 0.0528 x 24.3050  = 1.28 grams of Magnesium per Kg in seawater.
 

Offline miriam0920

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« Reply #23 on: 15/09/2008 01:45:19 »
Yes, I know I had to convert kilograms to centimeters equivalent to millilitres and by that liters, by cancelation units.

 

Offline RD

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« Reply #24 on: 15/09/2008 02:39:48 »
If total volume of seawater is 1.5 x 10(power of 21)L, and  Mg found in seawater is 1.3g of Mg for every kilogram of seawater, calcite how many Liters of seawater is need to extract 8.0 x 10 (power of 4) tons of Mg?

If you have posted the question correctly then the examiner wants you to convert kilograms seawater into litres of seawater. To do this you must have a value for the density of seawater, it should be given in the question or the exam paper or a data book/sheet you have.
    I have used a commonly used value for the density of seawater which is 1025 Kg m3 ,
 this means 1Kg of seawater occupies 1000/1025 = 0.975 litres .
i.e. to convert from kilograms of seawater to litres of seawater multiply by 0.975   
« Last Edit: 15/09/2008 02:43:50 by RD »
 

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