Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
ask your lecturer for help!
He isn't replying to emails and I'm not in Cork!!!!!!!!!!
f**k off illdecide how is it showing off when I can't do the bloody thing
will get ya the answer after 6, my wife is not home till then ;)
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
I generally find a calculator is good at Maths, ask one of those.
Quote from: thebigfella on April 27, 2009, 05:37:31 PM
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
I generally find a calculator is good at Maths, ask one of those.
Oh silly me...
you think I haven't been putting those numbers into the calculator to try and get that number?????!!!!!!!!!!!!!!!!!!!!!
Sweet feck how the hell wud any of us know that, more to the point why wud any of us want to know the answer to that
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
I'm sure someone will take up the challenge but failing that I'll get you an answer tomorrow as I have about 40 analytical chemists at my disposal!
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
Our calculations show that the bird is greater than or equal to the word.
I'm asking for help how is that showing off
if you don't like it then don't look at it
Quote from: corn02 on April 27, 2009, 07:09:37 PM
Our calculations show that the bird is greater than or equal to the word.
:D
Our calculations show that the bird is greater than or equal to the word.
[/quote]
:D :D :D
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
Fairly straightforward. The low-temperature optical properties of submonolayer CdSe/ZnSe nanostructures have been investigated with nominal layer thickness of 0.15 ML and 0.58 ML CdSe. In photoluminescence excitation spectroscopy (PLE) we observe more than 10 equidistant peaks separated by the ZnSe LO-phonon energy even at energies high above the ZnSe-band edge. The peak heights and line shapes are extremely sensitive to the monolayer coverage, have strongly different intensities in emission and PLE and deviate from the Poissonian intensity distribution expected within the concept of a constant, material specific Huang-Rhys parameter S. A bound polaron model is proposed to explain the optical properties of quantum structures for which the approach of a pure electronic quantum confinement is not appropriate. Intrinsic coherence of joint exciton-phonon modes is very promising, e.g. for application in quantum information processing or coherent population transfer.
He's a teacher so it must be right ;)
Where would we be without O'Neill!! :D
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
DG,
the 2.794 ml is the volume part of the answer, do you have all the right data?
Try www.mychemistrytutor.com
Shane there is at least two school boy errors there. I am not going to waste my time pointing them out to you.
Then basically it works out as
MnO4- : Fe2+
1:5
Where did you get the above information from?
Quote from: DrinkingHarp on April 27, 2009, 08:15:50 PM
Quote from: downgirl on April 27, 2009, 05:21:08 PM
Ok it is chemistry based but it is analytical so I'm hoping someone who is good at maths can help me out...I have a sample answer for it but I don't know where the figures are coming from...
This is the question:
1g of a sample of iron ore was dissolved in H2SO4 in a titration vessel yielding Fe2+. This solution was titrated using 0.1M KMnO4.
The analysis was repeated five times with the following titration volumes:
5.0, 4.8, 5.1, 5.2 and 5.3ml.
Calculate the average iron ore content in the ore as a mass percentage.
Aw(Fe) = 55.84 gmol-1
Then basically it works out as
MnO4- : Fe2+
1:5
then it has written down
5n (MnO4-) = n Fe2+
Ans x(%)Fe = 2.794 ml V KMnO4
How do you work out the 2.794 part?????
DG,
the 2.794 ml is the volume part of the answer, do you have all the right data?
Try www.mychemistrytutor.com
Thanks for the link DH gonna give it a try now...the question is exactly as I have put it up, I don't think I have enough info to answer it as I don't have the molarity of the iron solution...
Tony is hopefully gonna get me sorted out...I could kiss him!!
Quote from: Puckoon on April 27, 2009, 08:49:16 PM
Then basically it works out as
MnO4- : Fe2+
1:5
Where did you get the above information from?
equation for the rxn
MnO4- + 5Fe2+ + 8H+ --> Mn2+ + 5Fe3+ + 4H20
1 : 5
surely tony wouldn't be that brave
f**k off illdecide how is it showing off when I can't do the bloody thing
[/quote]
No need to be rude now i was only busting your chaps
The problem stated is simple and has nothing inherent that can be perceived as difficult. The fact is that H2SO4 is sulphuric acid thus it will react as determined by the equation. However, what should be borne in mind is that anyone who sat on their pc and tried to determine the answer to this really has only two options at stake - Chemistry and Masturbation. So. if you have been engrossed by the equation - you really need to get a girlfriend. To put it in terms of Physics - you have a tendency to indulge in rapid friction accumulation that will result in an emission over the digits! Horses for Courses! :P
Quote from: cville on April 27, 2009, 11:15:31 PM
The problem stated is simple and has nothing inherent that can be perceived as difficult. The fact is that H2SO4 is sulphuric acid thus it will react as determined by the equation. However, what should be borne in mind is that anyone who sat on their pc and tried to determine the answer to this really has only two options at stake - Chemistry and Masturbation. So. if you have been engrossed by the equation - you really need to get a girlfriend. To put it in terms of Physics - you have a tendency to indulge in rapid friction accumulation that will result in an emission over the digits! Horses for Courses! :P
Wonderful analytical powers there cville. Did you notice the 'girl' part of the name Downgirl?