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Question:If a sphere of volume 381.704m^3 is filled with Helium at 27 degrees C and a pressure of 200 KPa how do you calculate the mole number and mass of Helium in the balloon.
Answers:381.704 m^3 = 381704 dm^3 T = 27 + 273 = 300 K p = 200000/ 101325=1.97 atm n = pV/RT = 1.97 x 381704/ 0.08206 x 300=3.06 x 10^4 =moles He mass He = 3.06 x 10^4 mol x 4.00 g/mol=1.22 x 10^5 g
Answers:381.704 m^3 = 381704 dm^3 T = 27 + 273 = 300 K p = 200000/ 101325=1.97 atm n = pV/RT = 1.97 x 381704/ 0.08206 x 300=3.06 x 10^4 =moles He mass He = 3.06 x 10^4 mol x 4.00 g/mol=1.22 x 10^5 g
Question:Naturally occurring antimony has a molar mass of 121.84 g/mol and contains only two isotopes. One is 121Sb, which is 57.3% abundant. What is the mass number of the other isotope?
I'm not sure how to go about this. We haven't done it in class so I can't follow an example, so if someone could explain what to do that would be really appreciated. Thanks.
Answers:ok so the 121.84 is a average of teh precentage by weight of all isotopes 121Sb has 51 protons and 70 neutrons giving it 121g multiple this by .573 to get the weight it is contributing to the 121.84 then subtract this from the 121.84 and divide by 42.7 (the other part of teh precent 10057.3) this will be the weight so: 121*.573=69.33 g 121.8469.33=52.52g 52.52/.427=122.97g so rounded it would be 123 grams, have 51 protons and 72 neutrons hope this helped ~~~?!?
Answers:ok so the 121.84 is a average of teh precentage by weight of all isotopes 121Sb has 51 protons and 70 neutrons giving it 121g multiple this by .573 to get the weight it is contributing to the 121.84 then subtract this from the 121.84 and divide by 42.7 (the other part of teh precent 10057.3) this will be the weight so: 121*.573=69.33 g 121.8469.33=52.52g 52.52/.427=122.97g so rounded it would be 123 grams, have 51 protons and 72 neutrons hope this helped ~~~?!?
Question:a. Ascorbic acid ( C6H8O6)
b. sulfuric acid (H2SO4)
c. silver nitrate (AgNO3)
d. Saccharin (C7H5NO3S)
Answers:Calculating molar mass of any molecule becomes easy if you know the molecular formula i.e. the atoms present, number of those atoms in the moleucle and their respective atomic mass. The elements and their respective atomic masses appearing in the molecules you asked are as follows: Atom Atomic mass Carbon 12 Hydrogen 1 Oxygen 16 Sulphur 32 Silver 107 Nitrogen 14 Molar mass of ascorbic acid (C6H8O6) is: = (6 X Atomic mass of Carbon) + (8 X Atomic mass of Hydrogen) + (6 X Atomic mass of Oxygen) = (6 X 12) + (8 X 1) + (6 X 16) = 72 + 8 + 96 = 176 grams per mole Molar mass of Sulfuric acid (H2SO4) (note that here there is 1 sulphur atom in 1 molecule of H2SO4) is: = (2 X Atomic mass of Hydrogen) + (1 X Atomic mass of Sulphur) + (4 X Atomic mass of Oxygen) = (2 X 1) + (1 X 32) + (4 X 16) = 2 + 32 + 64 = 98 grams per mole Molar mass of Silver nitrate (AgNO3) is: = (1 X Atomic mass of Silver) + (1 X Atomic mass of Nitrogen) + (3 X Atomic mass of Oxygen) = (1 X 107) + (1 X 14) + (3 X 16) = 107 + 14 + 48 = 169 grams per mole Molar mass of Saccharin (C7H5NO3S) is: = (7 X Atomic mass of Carbon) + (5 X Atomic mass of Hydrogen) + (1 X Atomic mass of Nitrogen) + (3 X Atomic mass of Oxygen) + (1 X Atomic mass of Sulphur) = (7 X 12) + (5 X 1) + (1 X 14) + (3 X 16) + (1 X 32) = 84 + 5 + 14 + 48 + 32 = 183 grams per mole I hope you've been satisfied by the answer and now know how to calculate the molar mass of any molecule
Answers:Calculating molar mass of any molecule becomes easy if you know the molecular formula i.e. the atoms present, number of those atoms in the moleucle and their respective atomic mass. The elements and their respective atomic masses appearing in the molecules you asked are as follows: Atom Atomic mass Carbon 12 Hydrogen 1 Oxygen 16 Sulphur 32 Silver 107 Nitrogen 14 Molar mass of ascorbic acid (C6H8O6) is: = (6 X Atomic mass of Carbon) + (8 X Atomic mass of Hydrogen) + (6 X Atomic mass of Oxygen) = (6 X 12) + (8 X 1) + (6 X 16) = 72 + 8 + 96 = 176 grams per mole Molar mass of Sulfuric acid (H2SO4) (note that here there is 1 sulphur atom in 1 molecule of H2SO4) is: = (2 X Atomic mass of Hydrogen) + (1 X Atomic mass of Sulphur) + (4 X Atomic mass of Oxygen) = (2 X 1) + (1 X 32) + (4 X 16) = 2 + 32 + 64 = 98 grams per mole Molar mass of Silver nitrate (AgNO3) is: = (1 X Atomic mass of Silver) + (1 X Atomic mass of Nitrogen) + (3 X Atomic mass of Oxygen) = (1 X 107) + (1 X 14) + (3 X 16) = 107 + 14 + 48 = 169 grams per mole Molar mass of Saccharin (C7H5NO3S) is: = (7 X Atomic mass of Carbon) + (5 X Atomic mass of Hydrogen) + (1 X Atomic mass of Nitrogen) + (3 X Atomic mass of Oxygen) + (1 X Atomic mass of Sulphur) = (7 X 12) + (5 X 1) + (1 X 14) + (3 X 16) + (1 X 32) = 84 + 5 + 14 + 48 + 32 = 183 grams per mole I hope you've been satisfied by the answer and now know how to calculate the molar mass of any molecule
Question:A 0.391 m solution of the solute hexane dissolved in the solvent benzene is available. Calculate the mass (g) of the solution that must be taken to obtain 247 g of CH3(CH2)4CH3.
and
Calculate the mass (g) of the solute C6H6 and the mass (g) of the solvent tetrahydrofuran that should be added to prepare 1.63 kg of a solution that is 1.42 m.
Can somebody please explain how to solve these problems? Your help is much appreciated. :)
Answers:(247 g hexane)(mol hexane / 86.175 g hexane)(kg benzene / 0.391 mol hexane)(1000 g/kg) = 7.33e3 g benzene 7.33e3 g + 247 g = 7.58e3 g solution  (1.42 mol benzene / kg THF)(78.1118 g benzene / mol benzene) = 111 g benzene / kg THF 111 g + 1000 g = 1111 g solution (111/1111)(1630 g) = 163 g solute (1000/1111)(1630 g) = 1.47e3 g solvent
Answers:(247 g hexane)(mol hexane / 86.175 g hexane)(kg benzene / 0.391 mol hexane)(1000 g/kg) = 7.33e3 g benzene 7.33e3 g + 247 g = 7.58e3 g solution  (1.42 mol benzene / kg THF)(78.1118 g benzene / mol benzene) = 111 g benzene / kg THF 111 g + 1000 g = 1111 g solution (111/1111)(1630 g) = 163 g solute (1000/1111)(1630 g) = 1.47e3 g solvent
From Youtube
Video 3 Atomic Mass Calculation Pennium Lab :This video shows how to calculate weighted atomic masses from the mass numbers and abundances of the isotopes of an element.
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