how do i calculate serum albumin using--C1V1=C2V...\nhow do i calculate serum albumin using--C1V1=C2V...\nstock serum albumin contains 10 mg per cm3.\npipete 1cm3 serum albumin to tube 2, 2cm3 to tube 3, 3cm3 to tube 4, 4cm3 to tube 5, 5cm3 to tube 6---- all tube mixed with distilled water up to 10 cm3,......
200871
Concentation 1 x volume 1 = conc 2 x volume 2.\n\nYoure are probably using it to find total blood volume with a known volume of 'marker' albumin.  \nThe marker concentration x its volume = the concentration after dilution x the amount of diluent(the blood volume)
388178
You need to calculate the molecular weight of fluoxymesterone first, which I will label as x here.  And since all of the following answers are based off of this number, I will use A, B, C for anything that has an answer dependent on x or a previous answer.\n\ncalculate the moles of fluoxymesterone:  17mg x 1g/1000mg x 1mol/x g = A mol of fluoxymesterone.\n\ncalculate the molarity of the stock solution: A mol fluoxymesterone / 0.5 L = B M.\n\ncalculate how much fluoxymesterone is in your aliquot: 100uL x 1L/1000000 uL = 0.00001 L x B mol/L = C mol of fluoxymesterone present in your aliquot.\n\ncalculate the final solution in terms of molarity: C mol/0.1L = D M.
183108
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
414257
118 gram of C3H8 is 118/(3*12+8) = 2.682 mol.\nThe reaction formula is C3H8 + 5 O2 -> 3 CO2 + 4 H2O, therefore 5*2.682 = 13.4 mol O2 is needed.\nAt STP this is 13.4 * 22.4 = 300 liter.\nSTP corresponds to 273 K and 101.3 kPa (760 torr).\nAs PV/T is constant, we can write\nV = (P_STP/P) (T/T_STP) V_STP = (100/101.3) (302/273) 300 =328 liter = 0.328 m^3.
444652
Q10 is a ratio:\n\nHow fast an enzyme works at (T+10) C / How fast the enzyme works at T C\n\nIn other words, by how much does the rate of enzyme action increase when the temperature is raised by 10oC.\n\nMost school texts give it as 2, but it's never been 2 in any experiments my classes have done!
448384
C1V1 =  C2V2     thereafore C2  =  (C1*V1)/V2\n\nC1=3.58 * 10E-3  M\nV1=5.0 * 10E-3 dm3\n\nC2 = x M\nV2 = 2.5 * 10E-2\n\n\nC2 = (3.58*10E-3      X     5.0*10E-3) / 2.5*10E-2\n\nFinish off
549518
Use the formula \nCstock*Vstock= Cdil* Vdil=>\nVstock= (Cdil/Cstock) *Vdil=\n=(0.1/1)*100 =10 ml\n\nSo you need 10ml of stock solution. Put it in a volumetric cylinder or flask of 100ml, add water up to 100ml and mix.\n\nMixing 10ml stock and 90ml solvent is bad practice for making solutions. It is an approximation used only when you are handling very small volumes (<2ml) because there are no volumetric cylinders or flasks for such volumes.
425299
(1) Find the empirical formula.  Assuming a 100 gram sample,\n\n40.00 g C x (1 mol C/12.011 g C) = 3.33 mol C\n6.71 g H x (1 mol H/1.0079 g H) = 6.66 mol H\n53.28 g O x (1 mol O/16.00 g O) = 3.33 mol O\n\nSo the ratio of C:H:O is 1:2:1, and the Empirical Formula is C1H2O1.\n\n(2) Now the molecular formula of something is some multiple of the empirical formula, or (C1H2O1) x z.  We have to figure out what this z is.\n\nIf it's a gas, and the data given are: \nP = 2931.6 mmHg x (1 atm/760 mmHg) = 3.86 atm\nV = 189.0 mL = 0.1890 L\nT = 23.0C = 296 K\nR = 0.0821 atm-L/mol-K\nso using the Ideal Gas Law, PV=nRT, n  = 0.0300 mol\n\nThe molar mass of the compound is mass/moles = m/n = 1.80 g/0.0300 mol = 60.0 g/mol.\n\nThe Empirical Formula is C1H2O1; one empirical formula unit has a mass of (1 x 12.011) + (2 x 1.0079) + (1 x 16.00) = 30.03; if the molecule weighs 60.0 g/mol, then the z is 2 (in other words, it requires two empirical formula units to make up the molecule) nad the molecular formula is C2H4O2.
481578
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
488243
Place a 2 in front of the HCl to balance the equation. Then you can see that for one mole of calcium carbonate, you get 1 mole of carbon dioxide.
240991
Just to warn you, if you read the answers you will never benefeit from this problem as you normally would. And learning this material will be more difficult as time continues. Years from now you will wish you tried harder in this class but I am done... heres a good set of ansewers for you to learn from.\n\n\n\n4. for an ideal gas PV=nRT\n    \nmm = molar mass = m/n \nwhere m = mass and n = moles\n\n   therefore n = m/mm\n\n and PV=(m/mm)RT\n     and mm=mRT/PV\n          \nwhere R is the ideal gas constant, T is temp, P is pressure, and V is volume. Plug your stuff in and make sure to cancel out units.\n\n5. Using Dalton's Law of Partial Pressures\nwe know that for P=Pressure\n P(He) + P(O2) = Total Pressure\nP=nRT/V with only n being different between gases for P. Therefore,\n\n     (RT/V)[ n(He) + n(O2)] = Total Pressure.\n\nusing the relationship n=m/mm\n\n    (RT/V) [m(He)/mm(He)  + m(O2)/mm(O2)] = Total Pressure\n\nand you plug everything in and solve for T/V. And using this you can find the pressure of He alone. Let this ration T/V equal A for now.\n\n       Using PV=nRT,   P=mRT/mmV, or P=mRA/mm\n\n6. 6 d electrons. The correct answer also depends on the symmetry and ligands of the complex, but they could just be asking you to count electrons. In an octahedral low spin complex it would have 0 unpaired electrons. In an octahedral high spin complex it would have 1 unpaired electron. In a tetrahedral complex it would have 4 unpaired electrons. And there are about 20 more symmetries but these are the ones people talk about in beginning courses.\n\nps. you can do your own math.
524262
The dilution formula is:\n\n(M1 and V1 are the concentration and volume of the stock solution; M2 and V2 are the concentration and volume of final solution.  M1 and M2 should be in molarities; V1 and V2 should be liters, but it doesn't matter as long as they're the same units)\nM1V1 = M2V2\n(12.0 M)(V1) = (2.50 M)(250.0 mL)\nV1 = 52.1 mL\n\n(C)
473024
The product of a neutrilization reaction is always water and a salt, so use that as a guide.  For part 2, use the units as a guide.  Hint M=molarity=mols/L
277937
MOLARITY=(MOLECULAR WT/GIVEN WT)*(1000/VOLUME OF THE SOLUTION IN MILLIITERS)\n\nBUT WE KNOW THT M1V1=M2V2\nie 35*2.104=M2*825\nM2=0.0896M\nTHE AMOUNT OF WATER ADDED IS 825-35=790.
62742
C6H12O6 + 6O2->6CO2+6H2O\n\nOh you are right, I never did see you have a completely incorrect answer that magically changed to an answer with faint resemblence to mine! Man, I should wear glasses!
423042
Great problem!  OK, weigh the chalk before and after you write your name.  Assuming no chalk dust, then the difference in weight is the amount that it took.  Let's say that the amount is 1.0009.  The molecular weight of calcium carbonate is 100.09, so you've used (1.0009/100.09) = 0.01 moles of CaCO3.  Using Avogadro's number, this would be 0.01 moles x 6.023 x 10^23 molecules/mole or 6.023 x 10^21 molecules.  So you have that many calcium atoms, that many carbon atoms, and three times that many oxygen atoms.
117339
By serum protein electrophoresis, or by subtracting Serum albumin value from serum total protein value