Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: miriam0920 on 12/09/2008 21:48:59
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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.
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Miriam...
You're not asking us to do your homework for you, are you? [;)]
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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.
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I have the answer yet I want to learn how to calculate something like this.
I'm burn out.
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Where are my friends Bored Chemist and Lightarrow? Please help.
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Is "ton"
metric ton = tonne = 1000 kg
imperial ton = 1016 kg
short ton (USA) = 907 kg
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RD, in 1 ton = 2000 lbs so I guess I have to cancel lbs to gram which is 453.6 g/1 lbs.
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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.
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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.
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Initially I didn't factor-in the increased density of seawater.
The density of seawater varies 1020-1035Kgm3 (http://hypertextbook.com/facts/2002/EdwardLaValley.shtml), I used 1025 Kgm3, your question may use a different value.
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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.
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So that means that if I use your factor I might get it wrong?
1025 Kgm3 appears to be the most commonly used value (http://en.wikipedia.org/wiki/Seawater).
Your question or text book could have a figure anywhere between 1020 and 1035 Kgm3.
To obtain 8.0 x104 (short) tons (http://en.wikipedia.org/wiki/Short_ton) of Mg, i.e. 8.0 x104 x 907 Kg (http://en.wikipedia.org/wiki/Short_ton) = 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.
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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.
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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. :(
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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)
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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.
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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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Sorry, I forgot about the Americans using tons that don't weigh much.
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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?
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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.
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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?
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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].
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Here are the Wikipedia sources I have used...
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.
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Yes, I know I had to convert kilograms to centimeters equivalent to millilitres and by that liters, by cancelation units.
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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
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Okay, so I finally went to Chemistry class and the professor told me that the density of seawater was indeed needed as R/D stated. So the density of seawater was given in the problem prior to this one. You guys are really good! I mean, you also got the questions correct without mostly giving the density of seawater which was 1.03g/mL.
Just to let you know tomorrow is my big day, my first chemistry examen and my first lab examen. Thank you to all of you that helped me.
My heroes = RD, Bored Chemist, Lightarrow.
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the density of seawater which was 1.03g/mL.
Using 1.03g/ml for the density of seawater then multiply by 1/1.03 = 0.971 to convert kilograms of seawater to litres of seawater.
(ignore the conversion factor of 0.975 which I gave in my previous post which was based on density 1.025g/ml)