Qualification > Sciences
chem question
slvri:
to make the ratio a whole number......
1.33 isnt close enough to 1 to make it approximately equal to 1.........but if it were 1.03 i would have approximated it to 1
Ghost Of Highbury:
--- Quote from: slvri on September 16, 2009, 04:34:58 pm ---to make the ratio a whole number......
1.33 isnt close enough to 1 to make it approximately equal to 1.........but if it were 1.03 i would have approximated it to 1
--- End quote ---
ohh..okk..
Bani:
Hey guys!
Here is a question I need help on:
i)-Oceanographers studying plankton found that a sample of seawater contained 1.20 nanomol dm^-3 of chlorophyll, C55H77MgN4O5. (1 nanomol= 1*10^-9 mol)
What mass of magnesium would be present in 1.00 cm^3 of this sample of seawater? Give your answer to three significant figures.
ii)- X-ray diffraction can be used to locate atoms or ions in molecules like chlorophyll. X-rays are scattered by the electrons in atoms and ions. In chlorophyll the atoms of one of the elements still cannot be located with certainty by this technique.
Suggest which element is most difficult to locate? (i think its hydrogen....not sure)
thanks in advance!
nid404:
ok here's the answer
no on moles= 1.2X10^-9 in one dm^3
therefore mass of Mg= 1.2X10^-9 X 24(mass of one mole)/ 1000( 1dm^3=1000cm^3)
=2.88X10^-11 g
check if it's right plz
and for the second one
I too think it's hydrogen because it has the least no of electrons in it's atom...which makes it difficult to locate it....diffraction is not enough to locate it
mousa:
--- Quote from: nid404 on September 18, 2009, 01:38:37 pm ---ok here's the answer
no on moles= 1.2X10^-9 in one dm^3
therefore mass of Mg= 1.2X10^-9 X 24(mass of one mole)/ 1000( 1dm^3=1000cm^3)
=2.88X10^-11 g
check if it's right plz
and for the second one
I too think it's hydrogen because it has the least no of electrons in it's atom...which makes it difficult to locate it....diffraction is not enough to locate it
--- End quote ---
yup.. thats right,i got the same answerss
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