I'm working on a lab report for chemistry. I have to find the solubility of the solvents in oil and water. Can you help me please?!
201139
The Merck Index and the Handbook of Chemistry and Physics are good resources for this.
8563
If you know the compounds you tested you could always try a MSDS (Material saftey data sheet). This will tell you the lethal dose of the chemicals but you must be more specific than oil and grease. There are many different forms of petroleum and hydrocarbons.
217892
You need to know concentrations to determine the amount of substance that precipitates out by using solubility products. In this case the precipitate will be CaSO4, so the solubility product will be given by:\n\nk=[Ca(+2)]*[SO4(-2)]\n\nWhere k is the equilibrium constant (probably a given), and [Ca(+2)] & [SO4(-2)] are the concentrations of Ca(+2) and SO4(-2) in the solution. Any amounts greater than the equilbrium concentration will precipitate out as a solid.
525814
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
421793
The ingredient in the greater amount is called the solvent, so oil in water means there is more water. Water in oil, more oil.  An emulsion of oil and vinegar(5% acetic acid in water) is mayonnaise. Salad dressing that you have to shake up are emulsified for a short time. There are wall paints that are emulsions of oils and water. Whe you have a leaky head gasket in an car, oil gets into the radiator and makes a messy emulsion.
524390
2.5% of 150 grams is 3.75 grams (150 x 2.5/100)\n\nThe amount of aqueous solvent (water) you need is 150 grams minus 3.75 grams of solute.\n\nAnswer:  Add 3.75 grams KCl to 146.25 grams water
110366
The ratio of the concentration (in mole fractions) of the solute in one solvent when compared to a second solvent gives you the partition coefficient.\n\nIn your case the calculations are simplified since you are using equal amounts of solvent.  With a partition coefficient of 9, you will have 9 times as much solute in the CH2Cl2 as you will the water.  Therefore the CH2Cl2 will have 40.5 grams of solute and the water phase will have 4.5 grams of solute.
107966
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
99988
Answer was found in the Handbook of Chemistry and Physics\n\nKMn04 is only slightly solube and as you have aready found the\nsolubility constant.\n\nNaMnO4 is list in the hand book as Very Soluble both in cold and hot water, and did not give a constant. That means that a lot of NaMnO4 can be dissolve into water, somewhat like table salt, NaCl\n\nNaMnO4 is often found in liquid form at about 40% in water, and it is a dark purple liquid. you can get this data sheet \nby just type in Yahoo search "sodium permanganate solubility"\n\ngood luck\n\nDavid
488479
Insufficient information to answer...is it 10% w:w, or 10% w:v?\nIf 10% w:w then it's 10 g of benzoic acid per 100 gms of solvent\nif 10% w:v then its 10g of benzoic acid per 100 cm3 of solution\n\nMost %s are w:v so if it is, then that means that the solution is 100 g benzoic acid per dm3...since FWt of benzoic acid is 122g, then the solution is 100/122 mol per dm3 (sorry don't have calculator handy!)
472355
What you may have is an amide.  The carbonyl group can act as a lubricant while the amine portion increases it's water solubility. Cooling and lubrication when mixed with water.  Good combination!
425131
If by 'undissolved' you mean why are some salts insoluble in water, then I think we are in business.\n\nIn order for a salt to be soluble it has to be energetically favorable for the ions to separate. This requires a negative gibbs free energy. G = H - TS So you have to find the H and S values for the solid to aqueous solution and see if G will be negative for a given temperature.\n\nA method I recently learned to make insolube salts more soluble is to add other ions to the water...but you have to be careful to not add a common ion...or Le Chantliear will come into play.
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)
478652
The samples must be homogenously ground to enable the solvent to fully soak into the sample and extract the compound you want in question. Also there is a question of statical averages so a uniformed powder sample is the best for this process. If the sample is lumply or too big the solvent can't fully soak into it and wherefor not all of your compound in question is extracted by the process (Then the solvent fills up the soxhlet thumbie (sp)). If this happens you will get a incorrect value then you come to do the maths in determinating the amoult of the coumpound in question.
510709
There is a mnemonic to easily remember the basic tenet of solubility: "Like dissolves like.." Salt, NaCl, is an ionic compound. Water is a polar compound. That is why water dissolves salt, becuase they could both be considered polar, they have a positive and negative end. Butanol is non-polar. That is why it dissolves napthalene, which is also non-polar, better than water.
434614
1-hexanol should not be that much soluble in water because of the presence large non polar group (Six Carbon) and a small polar group (1 OH).  On the other hand, it should be more soluble in hexane because of almost similar non polarity.\n\nAs the number of carbons increase, the polarity of the water molecule has less and less effect and disperson forces become more dominent force of attraction.
93976
um i don't know if this site has it, but you might want to try chemguide.co.uk :) it's a pretty good site.