Jump to content

Pereira

Members
  • Posts

    5
  • Joined

  • Last visited

Posts posted by Pereira

  1. Hello,

    Any help will be appreciated.

     

    Q:Determine moles of dichromate ion that reacted with iron(II) in titration (use average volume).

     

     

    My A: 0.0119mol

     

    6 Fe2+ + Cr2O72- + 14 H3O+ 6 Fe3+ + 2 Cr3+ + 21 H2O

     

    4g-Fe2+

     

    4g*(1mol/558.845g/mol) = 0.0716 moles of Fe

     

    (6/0.0716)=1/x

    x=(0.0716)1/(6)

     

    x=0.0119mols of dichromate.

     

    My questions are:

    1) am I way off in my approach?

    2)Why does the question want me to use volume?

     

    It seems that it should not be this easy so I assume I am way off. Also the avg volume of Fe solution 6.75mL and the concentration was =0.102mo/L.

     

    ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

    Alright before I submit this thread to publish I tried another approach:

     

    molarity(Cr2O7 2-)= (molarity of Fe(II)/6)*(volume of Fe(II)/volume of Cr2O7)

    =[(0.102mol/L)/6)] * (0.00675L/0.005L)

    =0.017*1.35

    =0.2295

     

    m(Cr2O72-)=215.992 * .1L *0.2295

    m=4.95mols

     

    *this answer seems like I am more on the right track I think

     

    Are either of these very close. Any hints will be greatly appreciated.

    ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

  2. I am quite new to genetics as well so forgive me if I am wrong but you can cross heterozygous parents as well. Parent 1 would give you WG wG and parent 2 would give you wg haploid gametes. So when you plug those into the punnett square you should get 2 different diploid cells; this will be your f1 generation.

     

    Now if you cross f1 X f1 you take each of those diploid genotypes and work them into 4 different haploid gametes once again. Put those in a second punnett square and you will get the f2 generation phenotypes. It is common to get a 9:3:3:1 ratio but that is if both parents are homozygous. It will be a fraction of 16 (?/16). Hope this helps.

  3. Hello,

     

    I am a little confused about complete dominance. The question I am trying to solve is: A trihybrid pea plant having the genotype AaBbC(1)C(2) is self fertilized there is complete dominance at the A and B loci and co-dominance at the the C loci. What fraction of the progeny will be phenotypically different from the parents?(assume independant assortment).

     

    Now I know that the co-dominance C(1)C(2) will show through all the progeny and that A\- & B\- will also be phenotypically alike to the parents. So A\-B\-C(1)C(2) will be phenotypically the same. My question is if A\-bbC(1)C(2), aaB\-C(1)C(2) will cause the progeny to be phenotypically different. As well (correct me if I am wrong) the aabbC(1)C(2) if the homozygous will still be phenotypically different. I know this should be fairly easy but my text is really confusing me at the moment.

     

    So far I think that the ratio of differing progeny is 3/4. If I am wrong any help will be appreciated.

     

    Thank you.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.