my chemistry teacher assigned a lab where we gathered data then plotted them on a graph (showing the effect of concentration on rate of dissolution) & making a line of best fit.. she then said to find the "degree of fit." how is this calculated?
201085
See about 'regression'.
545811
moles = concentration* volume\n\nfind the moles of the substance with given concentration and volume\n\n\nwrite and balance the equation. determine the moles equivalence of every reactant then use the equation above
8480
There's a formula for volumes of gasses:\n\nPV = nRT.\n\nIf I remember right, P = pressure, V = volume, n = number of moles, T = temperature, and R = molar density, or something molar that is specific to the gas.
244937
Draw a scatter-plot. In short draw a graph of y against x.\nIf the graph show linear relationship, u may continue. If the graph is not linear(or curvilinear) ignore the questions as regression model is  only for linear model.\n\nSince u didn't include datas, i can only gives u equations\n\n\nQ1....\nthe mean for the number of years a teacher has taught \n    =  (∑ x)/n\n\n \nthe mean for the students' average course grade \n    =  (∑ y)/n\n\nQ2....\nthe variance for the number of years a teacher has taught \n     s²=[ ∑ (xi – mean for x value) ² ]/(n-1)\n\nthe variance for the students' average course grade\n     s²=[ ∑ (yi – mean for y value) ² ]/(n-1)\n\n\nQ3....\nthe standard deviation(s)\ns= √variance for the number of years a teacher has taught\n\ns= √variance for the students' average course grade\n\nQ4....\nthe z-score for the number of years a teacher has taught\n= (observed value of x-mean for x value)/ the standard deviation of x value\n\nthe z-score for the students' average course grade\n= (observed value of y-mean for y value)/ the standard deviation of y value.\n\nobserved value means u must calculate for each value one by one.\n\nQ5.....\nThe correlation coefficent(r)\n=1/(n-1) X  ∑(xi-mean value for x / s for x value)(yi- mean value for y)\n\ns stands for standard deviation and X is times and n is sample size\n\nThe equation looks quite tough isn't it? Normally we calculated r².\n\nr²= (SSTO-SSE)/SSTO\n\nwhere SSE = sum of squared errors\n          = ∑(yi-predicted value of y)²\n\nthe predicted value is base on the least square line( u have to draw a graph that has the sum of ( yi-predicted value of y) is smaller than any other line.) \n\n  \n     SSTO = sum of squares total\n          = ∑(yi-the mean for y value)²\n\nAfter u find r², find r by √r²\n      \nThe value of r always between -1 and +1\n#If u get value for r outside that range, means u have made some mistakes in calculation.\n#If u get r=1 or -1, it means that the relationship is perfectly linear and all data points fall on the same straight line.\n#r=+something means that when two varibles tend to increase together.(x increase, y increase)\n#r=0 means the best straight line is exactly horizontal\n\nQ6....\ni dont know about the coefficient of determination. Never heard such thing. Sorry.\n\nQ7....\nthe regression equation\n\ny-hat=b-nought  -  b-one   x\n\n  where    \nb-one = [∑(xi-the mean for x value)(yi-the mean for y value)]/  ∑(xi-the mean for x value)²\n\nb-nought = the mean for y value  -  b-one(the mean of x value)\n\nQ8....\n From y-hat=(b-nought)  -  (b-one) x\nJust simply substitude x=x1, therefore\n   predicted y = b-nought)  -  (b-one)x1\n\nQ9....\nI'm not sure about standard error of the estimate for predicting Y from X. But if if i'm not mistaken, the formula should be\n\nstandard error(s.e)= s/ [∑(yi-the mean for y value)²][b-one)\n\nwhere b-one=r(the standard deviation for x value)/(the standard deviation for y value)\n\nand s= √[SSE/(n-2)]\n\n\n\n\nGood luck
71443
Here are the equations you need to know:\n\nMolarity (M) = moles/Liters\n\nc=(t 1/2)(2k); \nor concentration  = half life (t 1/2) * 2 * rate constant (k)\n\nUsing this equation, you can solve for the rate constant, k.\n\nThen you can plug that value into the equation for a first order rate law:   \n\ncf=c1-kt; \nor final concentration = initial concentration - rate constant * time\n\nGood luck with the calculations!
630388
It is based on mean and standard deviations (statistical stuff). With just the information you gave, it is impossible to tell.\nWe would need to know the mean, standard devaition, and method of calculating scale...but then again if we had all this stuuf you would be able to asnwer your own question, right?
428496
Hi,\nThe osmotic pressure for dilute solutions will be given by the V'ant Hoff equation:\n\nOP = i*cRT\n\nWhere c is the concentration, R the gas constant and T the temperature in Kelvin.  i is the number of dissociated particles. For your ratio RT is constant for both solutions\n\nFor the organic sugar (let's label case 1) the osmotic pressure will be (in this case i=1 because the sugar is not dissociated):\n\nOP1 = 1 * 0.02 * RT\n\nFor case two the calcium (let's label case 2) the osmotic pressure will be (in this case i=3 because calcium chloride will dissociate in 3 particles)\n\nOP2 = 3 * 0.02 * RT\n\nSo:\n\nOP1 / OP2 = 1/3
114377
Unfortunately, this question can not be solved because it lacks pertinent information that is required to solve the problem.  But, I will lead you through the steps as if all the information were present.\n\nThe first step would be to determine how many moles of substance is present, the temperature(in Kelvins), and the volume(in L), using the R value of .0821.  This plugs into the equation PV=nRT.  Solve for P, which gives the total pressure of the system. \n\nThe next step is to determine the mole fraction for each of the substances.  This is accomplished by dividing the number of moles of the particular substance by the total moles used earlier.  This value is the mole fraction.\n\nFrom here, to find the partial pressure for the particular substance, multiply the mole fraction by the total pressure of the system.  This will give you the partial pressure for that substance.\n\nThen, using Dalton's Law of Partial Pressures, subtract the partial pressure value from the total pressure to determine the partial pressure of the other substance.  Now, the problem is solved.  Thus, the substance with the greatest mole fraction will have the greatest partial pressure.
96226
See comment above.\n\nYou do need to do several trials at various concentrations though. You can then determine it through a linear regression.\n\nI also suggest that you consult a book on Analytical chemistry, because Beer's law has limitations. For example, it loses linearity at higher concentrations.
350276
You can measure it in a number of ways depending on the protein.\n\n-- pH may vary directly with concentration\n-- optical rotation measurements may work for some\n-- adding a dye to the protein and reading on a colorimeter\n-- Another way would be to degrade or react the protein in solution and then to measure the byproducts of the degradation/reaction by something as simple as a titration
80080
Sorry, but I think you need to do this yourself.  Here are some guidelines though; give them a try.\n1.  Make a line drawing so you can see the info you're given and the path to the required answer.\n2.  Make a list of ALL the related formulae you were given.  Check each one against the given data; if you find one in which you have all the variables EXCEPT one, substitute and solve for the unknown.\n3.  Sometimes, you may have to use one formula to find a variable and THEN use that in ANOTHER formula to arrive at your ansver.\n\nGood luck!!
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)
100911
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
58840
You just draw a graph yourself.\n\n! ain't gonna graph for you. Rather ! love calculus.
443242
Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
468279
I dont understand what r u doing with "p*v/t"; but if assume ideal gas behavior, PV = nRT, so P = nRT/V [n is number of moles, R is ideal gas constant, T is temperature, V is volume of the flask, P is the pressure]
136809
You sure did pick a really long question for explaining.  And I'll have to argue that all of the pH's do not have correct sig figs but the numbers are right.\n\n(Sorry about the numbers in the tables but this place doesn't seem to allow better tables)\n\nA) 25.0mL of HAc(acetic acid) and 10.0mL of NaOH\n\n*This is a reaction so we have to use moles*\n\nHAc : 25.0mL * (0.100mol/L) * (1L/1000mL)= 0.00250 moles\n\nNaOH : 10.0mL * (0.125mol/L) * (1L/1000mL) = 0.00125 moles\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00125      0.00250        0\nChange:   -0.00125     -0.00125      +0.00125 \nEquilib:      0         0.00125       0.00125\n\nHAc and Ac-(NaAc breaks apart) for an equilibrium\n*Use Molarity when no reaction is taking place (best shown by when no volume is changing)*\n\nBoth HAc and Ac- have same molarity:\n\n(0.00125 moles / 35mL ) * (1000mL/1L) = 0.0357M HAc and Ac-\n\nUsing the equilibrium equation:\n\nEquation:  HAc   <--> Ac-     +  H+         Ka = 1.8 * 10^-5\nInitial:   0.0357M     0.0357M  \nChange:    -x          +x         x\nEquilib:   0.0357-x    0.0357+x    x\n\nKa=1.8*10^-5 = [Ac-][H+]/[HAc] = [0.00357+x][x]/[0.00357-x]\nKa=1.8*10^-5 = [0.00357][x]/[0.0357] usually good approx.\nKa = 1.8 * 10-5 = x = [H+]\n\npH = -log [H+] = -log [1.8*10^-5] = 4.745 (3 sig figs as only after the decimal counts for logs)\n\nB) 25.0mL HAc and 20.0mL NaOH\n\n*See A for HAc moles and double the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00250      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:      0            0          0.00250\n\nCalculate molarity of Ac-:\n\n(0.0250moles /45mL) * (1000mL/1L) = 0.0556M Ac-\n\nUsing the other equilibrium equation (Kb):\n\nEquation:  Ac- + H20   <--> HAc     +  OH-         Initial:   0.0556M   -        0\nChange:    -x        -        +x        + x\nEquilib:   0.0556-x  -       +x         + x\n\nThis gives us the Kb equation\nKb = ([HAc][OH-])/[Ac-] = with approx ([x][x])/[0.0556]\nso Kb = x^2 /0.0556\n\n**but we need to calculate Kb\n**Kw = 1*10^-14 at STP I'll assume this for this problem\n**Kw = KaKb so Kb = Kw/Ka\n**Kb = (1*10^-14)/(1.8*10^-5)= 5.56*10-10 \n\nso x = 5.56 * 10^-6 = [OH-]\n\npOH = -log [OH-] = 5.253\nPH = 14 - pOH (at STP)= 8.747\n\nC)25.0mL HAc and 30.0mL NaOH\n\n*See A for HAc moles and triple the NaOH for this trial*\n\n*This is a reaction so we have to use moles*\n\nEquation:   NaOH    +   HAc    -->   NaAc + H20\nInitial:   0.00375      0.00250        0\nChange:   -0.00250     -0.00250      +0.00250 \nEquilib:   0.00125            0       0.00250\n\nAfter the reaction there is some NaOH left over so it is the only thing that matters for the pH as it is a stong base.\n\nCalculate NaOH molarity\n(0.00125moles/55.0mL)*(1000mL/1L)= 0.0227M NaOH\n\nStrong Bases dissociate 100% so [OH-] = 0.0227M\n\npOH= -log[0.0227] = 1.644\npH = 14-pOH = 12.356\n\nThis should show how they calculted the half-neutralization point (A), the neutralization point (B) and after the neutralization point (C).\n\nIf you have questions on this, feel free to send me a message.  I taught this a couple weeks ago so I'm still pretty fresh on the subject.