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Question:Having major problems with this. Please help!
Answer these questions. Show your work.
16. What is the molar concentration (i.e., molarity) of sucrose (C12H22O11) if 150.0 g is dissolved in 250.0 mL of solution?
17. What is the molar concentration of methanol (CH3OH) if 125.0 mL is dissolved in enough water to make 15.0 L of solution? The density of methanol is 0.792 g/mL.
18. How many grams of NaOH are contained in 250.0 mL of a 0.400 M sodium hydroxide solution?
19. The drinking water standard for lead is 15 ppb (parts per billion). What is this in M?
20. What volume of 0.650 M sucrose must be diluted with water to prepare 250.0 mL of 0.423 M sucrose?
21. Which of the following solutions contains the largest concentration of chloride ions:
a. 100 mL of 0.30 M AlCl3
b. 50.0 mL of 0.60 M MgCl2
c. 200.0 mL of 0.40 M NaCl
Answers:16. Molar mass = 342 g/mol Moles = 150 / 342 = 0.439 M= 0.439 / 0.250 L = 1.75 M 17. Mass = 125 x 0.792 = 99 g Molar mass = 32 g/mol Moles = 99 / 32 = 3.09 M = 3.09 / 15 = 0.206 18. Moles = 0.250 x 0.400 = 0.10 Molar mass = 40 g/mol 40 x 0.1 = 4.0 g 19. 1 ppb = 1 mg / 1 m^3 = 1 mg / 1000 L 15 mg = 0.015 g Moles = 0.015 / 207.2 = 0.000072 M = 0.000072 / 1000 = 7.2 x 10^8 20. 0.25 x 0.423 = V x 0.65 V = 0.163 L = 163 mL 21. a. 0.3 x 0.1 = 0.03 moles AlCl3 => moles Cl = 0.09 b. 0.05 x 0.6 = 0.03 moles MgCl2 => moles Cl = 0.06 c. 0.2 x 0.4 = 0.08 moles NaCl => 0.08 moles Cl
Answers:16. Molar mass = 342 g/mol Moles = 150 / 342 = 0.439 M= 0.439 / 0.250 L = 1.75 M 17. Mass = 125 x 0.792 = 99 g Molar mass = 32 g/mol Moles = 99 / 32 = 3.09 M = 3.09 / 15 = 0.206 18. Moles = 0.250 x 0.400 = 0.10 Molar mass = 40 g/mol 40 x 0.1 = 4.0 g 19. 1 ppb = 1 mg / 1 m^3 = 1 mg / 1000 L 15 mg = 0.015 g Moles = 0.015 / 207.2 = 0.000072 M = 0.000072 / 1000 = 7.2 x 10^8 20. 0.25 x 0.423 = V x 0.65 V = 0.163 L = 163 mL 21. a. 0.3 x 0.1 = 0.03 moles AlCl3 => moles Cl = 0.09 b. 0.05 x 0.6 = 0.03 moles MgCl2 => moles Cl = 0.06 c. 0.2 x 0.4 = 0.08 moles NaCl => 0.08 moles Cl
Question:Please, can anybody tell me how to figure out the problems that my professor gave to my class to review for inorganic chemistry? He's giving us a test in a couple of days and I have no clue how to solve these problems. I've tried to find examples of how to solve them in my text book and on the web, but I'm not having any luck.
The following problems are:
1) Calculate the molarity of a solution that has been prepared disolving 10.1g of NaSO4 (99.8%) in enough water to obtain 1500ml of the solution.
2) If you mix 20ml of 0.20 M KCl, 30ml 0.60M K2SO4, and 20ml of 0.12M MgCl2, calculate the molarity of K+, Cl, and SO4. Calculate the pK and pCl.
3)Calculate the molarity of the solution that results from mixing 500ml of 3M H2SO4 with 1.5L 0.5M H2SO4.
Answers:Let's begin with a few simple definitions: a.) A mole is a grammolecularweight of a particular substance. This means you look up the atomic weight of each element (see Periodic Table) and multiply it by the number of times that element occurs in the molecule. Sum all the different element weights and you've got the total molecular weight of that substance. Then, if you use a balance to measure that amount of substance, in grams, you've got a mole of the stuff. b.) A one molar solution of the substance contains exactly one grammolecular weight of solute dissolved in enough solvent to total exactly one liter. A two molar solution would contain exactly two grammolecularweights of solute disolved in enough solvent to still total exactly one liter. Thus, a two molar solution is twice as concentrated as a one molar solution. Problem 1.) Begin by looking up the required atomic weights: Na = 22.989770 S = 32.065 O = 15.9994 From the formula we know that sodium and sulfer occur once and oxygen occurs four times in each molecule. Now we can calculate the molecular weight: 22.98977 + 32.065 + 4(15.9994) = 119.05237 That's the molecular weight of sodium sulfate. We'll save this result for later and move on to the next step. We have 10.1 g of 99.8% NaSO4, so we multiply: 10.1 x 0.998 = 10.0798 g of pure NaSO4. How many moles of pure NaSO4 do we now have? It's a simple ratio of how much we have to how much is one mole: 10.0798 g / 119.05237 g = 0.0846669 moles. Now, if we (hypothetically) mixed this much solute (NaSO4) with enough water to total one liter we would have a solution of approximately 0.085 molar concentration. However, we're going to mix up a total volume of 1500 ml (1.5 liters). This means our solution will be: 1000 / 1500 = 2/3 as concentrated. We can find the the concentration of the required amount of solution with one more multiplication: 0.66666666(0.0846669) = 0.0564445 molarity Problem 2.) As before, begin by copying all the required atomic weights from the Periodic Table: K = 39.0983 Cl = 35.453 S = 32.065 O = 15.9994 Mg = 24.3050 Note that we have three volumes of three concentrations of three different salts. We need to calculate the weight of each element in each of the solutions. We'll use those weights later. 0.020 l(0.20 M)39.0983 = 0.1563932 g K = 0.004 M K 0.020 l(0.20 M)35.453 = 0.141812 g Cl = 0.004 M Cl 0.030 l(0.60 M)39.0983(2) = 1.4075388 g K = 0.036 M K 0.030 l(0.60 M)32.065 = 0.57717 g S = 0.018 M S 0.030 l(0.60 M)15.9994(4) = 1.1519568 g O = 0.072 M O 0.020 l(0.12 M)24.3050 = 0.058332 g Mg = 0.0024 M Mg Total weight ( g ) of solute = 3.4932028 g Total weight ( M ) of solute = 0.1364 M Total volume ( ml ) of solution = 70 ml To calculate the molarity of K+ ions: 0.004 M + 0.036 M = 0.040 M K+ Since there are 70 ml of solution, the molarity of the solution is: 0.040 / 0.070 = 0.5714285 molar concentration. To calculate the molarity of CL ions: There are 0.004 M Cl As above, 0.004 / 0.070 = 0.0571428 molar concentration. To calculate the molarity of the sulfate ions we must first calculate the grammolecularweight of the SO4 ion: 32.065 + 4(15.9994) = 96.0626 g Now we total weight of the atoms used to make sulfate in our solution: 0.57717 g S + 1.1519568 g O = 1.7291268 g SO4 Dividing the amount of sulfate by the G.M.W. of sulfate tells how many moles of sulfate are in solution: 1.7291268 / 96.0626 = 0.018 moles of SO4 solute. One more division gives us tthe molarity of SO4: 0.018 M / 0.070 l = 0.2571428 molar concentration. Next we calculate the percentage of K and Cl. We know from above that the total weight of all solute atoms is 3.4932028 g. 0.1563932 + 1.4075388 = 1.563932 g K (1.563932 / 3.4932028)100 = 44.770719% K+ We know from above that there is 0.141812 g Cl, so: (0.141812 / 3.4932028)100 = 4.0596555% Cl [[Now that I've seen "Know It's" answer, I'm thinking he may be correct on the pK and pCl thing. Truthfully, I didn't know what the p stood for. I've assumed it was percentage and I could be wrong. Be careful here.]] Problem 3.) I do this type of problem by calculating the total moles of solute and the total volume of solution: (500 ml / 1000 ml)3 M/l = 1.500 moles of H2SO4 (1500 ml / 1000ml)0.50 m/l = 0.75 moles of H2SO4 Total moles of H2SO4 = 1.500 + 0.750 = 2.250 moles H2SO4 Total volume of solution = 0.50 l +1.50 l = 2 liters total solution Molarity of the solutionis calculated by: 2.250 M / 2 L = 1.125 Molar concentration.
Answers:Let's begin with a few simple definitions: a.) A mole is a grammolecularweight of a particular substance. This means you look up the atomic weight of each element (see Periodic Table) and multiply it by the number of times that element occurs in the molecule. Sum all the different element weights and you've got the total molecular weight of that substance. Then, if you use a balance to measure that amount of substance, in grams, you've got a mole of the stuff. b.) A one molar solution of the substance contains exactly one grammolecular weight of solute dissolved in enough solvent to total exactly one liter. A two molar solution would contain exactly two grammolecularweights of solute disolved in enough solvent to still total exactly one liter. Thus, a two molar solution is twice as concentrated as a one molar solution. Problem 1.) Begin by looking up the required atomic weights: Na = 22.989770 S = 32.065 O = 15.9994 From the formula we know that sodium and sulfer occur once and oxygen occurs four times in each molecule. Now we can calculate the molecular weight: 22.98977 + 32.065 + 4(15.9994) = 119.05237 That's the molecular weight of sodium sulfate. We'll save this result for later and move on to the next step. We have 10.1 g of 99.8% NaSO4, so we multiply: 10.1 x 0.998 = 10.0798 g of pure NaSO4. How many moles of pure NaSO4 do we now have? It's a simple ratio of how much we have to how much is one mole: 10.0798 g / 119.05237 g = 0.0846669 moles. Now, if we (hypothetically) mixed this much solute (NaSO4) with enough water to total one liter we would have a solution of approximately 0.085 molar concentration. However, we're going to mix up a total volume of 1500 ml (1.5 liters). This means our solution will be: 1000 / 1500 = 2/3 as concentrated. We can find the the concentration of the required amount of solution with one more multiplication: 0.66666666(0.0846669) = 0.0564445 molarity Problem 2.) As before, begin by copying all the required atomic weights from the Periodic Table: K = 39.0983 Cl = 35.453 S = 32.065 O = 15.9994 Mg = 24.3050 Note that we have three volumes of three concentrations of three different salts. We need to calculate the weight of each element in each of the solutions. We'll use those weights later. 0.020 l(0.20 M)39.0983 = 0.1563932 g K = 0.004 M K 0.020 l(0.20 M)35.453 = 0.141812 g Cl = 0.004 M Cl 0.030 l(0.60 M)39.0983(2) = 1.4075388 g K = 0.036 M K 0.030 l(0.60 M)32.065 = 0.57717 g S = 0.018 M S 0.030 l(0.60 M)15.9994(4) = 1.1519568 g O = 0.072 M O 0.020 l(0.12 M)24.3050 = 0.058332 g Mg = 0.0024 M Mg Total weight ( g ) of solute = 3.4932028 g Total weight ( M ) of solute = 0.1364 M Total volume ( ml ) of solution = 70 ml To calculate the molarity of K+ ions: 0.004 M + 0.036 M = 0.040 M K+ Since there are 70 ml of solution, the molarity of the solution is: 0.040 / 0.070 = 0.5714285 molar concentration. To calculate the molarity of CL ions: There are 0.004 M Cl As above, 0.004 / 0.070 = 0.0571428 molar concentration. To calculate the molarity of the sulfate ions we must first calculate the grammolecularweight of the SO4 ion: 32.065 + 4(15.9994) = 96.0626 g Now we total weight of the atoms used to make sulfate in our solution: 0.57717 g S + 1.1519568 g O = 1.7291268 g SO4 Dividing the amount of sulfate by the G.M.W. of sulfate tells how many moles of sulfate are in solution: 1.7291268 / 96.0626 = 0.018 moles of SO4 solute. One more division gives us tthe molarity of SO4: 0.018 M / 0.070 l = 0.2571428 molar concentration. Next we calculate the percentage of K and Cl. We know from above that the total weight of all solute atoms is 3.4932028 g. 0.1563932 + 1.4075388 = 1.563932 g K (1.563932 / 3.4932028)100 = 44.770719% K+ We know from above that there is 0.141812 g Cl, so: (0.141812 / 3.4932028)100 = 4.0596555% Cl [[Now that I've seen "Know It's" answer, I'm thinking he may be correct on the pK and pCl thing. Truthfully, I didn't know what the p stood for. I've assumed it was percentage and I could be wrong. Be careful here.]] Problem 3.) I do this type of problem by calculating the total moles of solute and the total volume of solution: (500 ml / 1000 ml)3 M/l = 1.500 moles of H2SO4 (1500 ml / 1000ml)0.50 m/l = 0.75 moles of H2SO4 Total moles of H2SO4 = 1.500 + 0.750 = 2.250 moles H2SO4 Total volume of solution = 0.50 l +1.50 l = 2 liters total solution Molarity of the solutionis calculated by: 2.250 M / 2 L = 1.125 Molar concentration.
Question:Please, can anybody tell me how to figure out the problems that my teacher gave to my class to review for an AP chemistry Test shes giving us a in a couple of days and I have no clue how to solve these problems.
The following problems are:
1) A solution of hydrogen peroxide is 30% H2O2 by mass and has a density of 1.11 g/cm3. What is the molarity, if even possible
2) Calculate the molarity of C2H5OH in a water solution that is prepared by mixing 50 ml of C2H5OH with 100 ml of Hs0 at 20 degrees Celsius. The density of the C2H5OH is 0.789 g/ml at 20 degrees Celsius.
3)What is the molarity of a solution of 50 g of propanol (CH3CH2CH2OH) in 152 ml water, if the density of water is 1.0 g/ml.
4). A solution containing 296.6g of Mg(NO3)2 per liter has a density of 1.114 g/ml. The molarity of the solution is what?
Answers:Molarity = moles per liter of solution 1) 1 L of 30% H2O2 weighs 1110 gram (1.11 g/cm^3 x 1000 cm^3 = 1110 g) 30% of 1110 g = 333 grams 333 g / 34 g/mol = 9.79 moles 30% H2O2 is 9.79 molar 2) 50 mL x 0.789 g/mL = 39.45 g 50 mL + 100 mL = 150 mL 39.45 g / 46.07 g/mol = 0.856 mol 0.856 mol / 0.150 mL = 5.71 molar Now yopu should have a "clue" as to how you can solve #3 and #4
Answers:Molarity = moles per liter of solution 1) 1 L of 30% H2O2 weighs 1110 gram (1.11 g/cm^3 x 1000 cm^3 = 1110 g) 30% of 1110 g = 333 grams 333 g / 34 g/mol = 9.79 moles 30% H2O2 is 9.79 molar 2) 50 mL x 0.789 g/mL = 39.45 g 50 mL + 100 mL = 150 mL 39.45 g / 46.07 g/mol = 0.856 mol 0.856 mol / 0.150 mL = 5.71 molar Now yopu should have a "clue" as to how you can solve #3 and #4
Question:I am given a pH of 4.72 +/ .02, using CH3COONa to CH3COOH
I got as far as [CH3COONa]= .912 * [CH3COOH]
Is .912 the answer? I have only done these sorts of problems for acid to base, in which I know you take the reciprocal, 1/.912. How do I do it with this though? Thanks
Answers:pH = pKa + log [CH3COO]/ [CH3COOH] pKa = 4.74 4.72  4.74 =  0.02 10^0.02 =0.955 = [CH3COO]/ [CH3COOH]
Answers:pH = pKa + log [CH3COO]/ [CH3COOH] pKa = 4.74 4.72  4.74 =  0.02 10^0.02 =0.955 = [CH3COO]/ [CH3COOH]