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Question:while at a store, you bought a box of cookies. b/c u were intrigued by watching this video u decided that u would like to find out the density of this box of cookies. the weight of the box of cookies is 8oz. or 226.8 g. the box is 20 cm*8cm*14cm What is the density. please show me how u did this. IM REALLY CONFUSED!!! thank u! i just couldnt remember what to do and need help on that one.
Answers:D = M/V (Density = Mass/Volume) Volume = l x w x h V = (20)(8)(14) So... D = 226.8 / (20)(8)(14) Hope this helped!
Answers:D = M/V (Density = Mass/Volume) Volume = l x w x h V = (20)(8)(14) So... D = 226.8 / (20)(8)(14) Hope this helped!
Question:At low temperatures, xenon freezes to a solid having a density of 5.741 g/cm3. Below 84 K, argon also freezes to a solid that has a similar structure but different volume and mass. If the volume of each unit cell in argon is 82 percent of the xenon unit cell, what is the predicted density of solid argon?
Answers:5.741 g/cm^3 *(.82)^3
Answers:5.741 g/cm^3 *(.82)^3
Question:An electric device delivers a current of 6 A for 10 seconds. How many electrons flow through this device?
I thought I need to use the equation I = delta Q / delta t but the answer is supposed to be 3.7x10^20 and I can't figure out how to get that answer.
Answers:The electronic charge constant (charge per electron) = 1.602 x 10^19 C The reciprocal of this number tells us how many electrons make 1 coulomb of charge. This is 6,242 x 10^18 electrons per coulomb. 1 Amp of current is 1 coulomb of charge per second So 6A for 10 seconds is 60 coulombs if charge transferred. 60 coulombs = 6.242 x 10^18 x 60 electrons = 3.745 x 10^20 electrons
Answers:The electronic charge constant (charge per electron) = 1.602 x 10^19 C The reciprocal of this number tells us how many electrons make 1 coulomb of charge. This is 6,242 x 10^18 electrons per coulomb. 1 Amp of current is 1 coulomb of charge per second So 6A for 10 seconds is 60 coulombs if charge transferred. 60 coulombs = 6.242 x 10^18 x 60 electrons = 3.745 x 10^20 electrons
Question:The ratio of the size of whole atom to its nucleus is 1 mm to 100 meters or about 1/1000. The mass of one proton/neutron is 1836 times heavier than electron. So the ratio would still be same even if I add a mass of one proton to nucleus for the compensation of all electrons of element of highest electronic configuration. Ignore the packing factor and compressing of nucleus in order to make it more dense.
So now density of an atom = Mass /volume.
Density= mass of all proton, neutron and one extra proton which was added for the compensation of electrons/ volume of whole atom
The figure is too small as 99.009% volume of the atom is empty .Now with the help of Avogadro s numbers we can calculate the number of atoms in a mass of 1 kg of any element say aluminum. So the theoretical density of 1 kg of Al will be 0.001% of its real desity as 99.009% volume of 1 kg of mass is empty.
So am I a crackpot to post this question??
Answers:It is perhaps useful to separate the density of the 'atom' from the density of its 'nucleus'! As you so rightly point out, the atom with its electron cloud and tiny nucleus is mostly empty space. Thus any attempt to calculate the density of the atom in terms of its overall volume and mass will run into problems. if, however, a nuclear model such as Bohr's Liquid Drop Model is used then we can calculate the mass and radius of the nucleus from:  The liquid drop model of the nucleus (Bohr 1936) gives the mass of the nucleus 'M' as:  M = Z.M(protons) +(A  Z).M(neutrons)  BE/c Where 'Z' is the proton number and 'A' the atomic number with 'BE' as the binding energy given by:  BE = av.A  4.ac.(Z.(Z1)/A^(1/3))  as.A^(2/3) ar.((A  2.Z) /A) + E Where the coefficients have values:  av = 14.0 Mev, ac =0.146 Mev ar = 19.4 Mev, as = 13.1 Mev E varies as follows:  A..........Z..........N..........E even....even....even....+ /2A odd.....even....odd......0 odd.....odd......even....0 odd.....odd......odd..... /2A And = 270 Mev. This mass equation is known as the 'semiempirical mass formula' and was derived by considering the nucleus as a liquid drop, with volume energy (the av term), Coulomb energy (the ac term), surface energy (the as term), a symmetry effect due to the neutron excess (the ar term). It works with medium and heavy nuclei (large numbers of nucleons) but does not work well with small or light nuclei (few nucleons). The mean radius of the nucleus is given, by scattering experiments, to be:  .............................. R(e) = (1.25 0.02).A x 10 m Converting the radius value into a volume and dividing into the calculated mass (see above) gives most nuclei as having a density of about 10 kg/m (10^18 if the superscript is not clear). Hence, returning to the atomic radius and Avogadro's number it is possible to use the nuclear mass and atomic volume to find a reasonable estimate of the material's average density.
Answers:It is perhaps useful to separate the density of the 'atom' from the density of its 'nucleus'! As you so rightly point out, the atom with its electron cloud and tiny nucleus is mostly empty space. Thus any attempt to calculate the density of the atom in terms of its overall volume and mass will run into problems. if, however, a nuclear model such as Bohr's Liquid Drop Model is used then we can calculate the mass and radius of the nucleus from:  The liquid drop model of the nucleus (Bohr 1936) gives the mass of the nucleus 'M' as:  M = Z.M(protons) +(A  Z).M(neutrons)  BE/c Where 'Z' is the proton number and 'A' the atomic number with 'BE' as the binding energy given by:  BE = av.A  4.ac.(Z.(Z1)/A^(1/3))  as.A^(2/3) ar.((A  2.Z) /A) + E Where the coefficients have values:  av = 14.0 Mev, ac =0.146 Mev ar = 19.4 Mev, as = 13.1 Mev E varies as follows:  A..........Z..........N..........E even....even....even....+ /2A odd.....even....odd......0 odd.....odd......even....0 odd.....odd......odd..... /2A And = 270 Mev. This mass equation is known as the 'semiempirical mass formula' and was derived by considering the nucleus as a liquid drop, with volume energy (the av term), Coulomb energy (the ac term), surface energy (the as term), a symmetry effect due to the neutron excess (the ar term). It works with medium and heavy nuclei (large numbers of nucleons) but does not work well with small or light nuclei (few nucleons). The mean radius of the nucleus is given, by scattering experiments, to be:  .............................. R(e) = (1.25 0.02).A x 10 m Converting the radius value into a volume and dividing into the calculated mass (see above) gives most nuclei as having a density of about 10 kg/m (10^18 if the superscript is not clear). Hence, returning to the atomic radius and Avogadro's number it is possible to use the nuclear mass and atomic volume to find a reasonable estimate of the material's average density.
From Youtube
Ground School: How to Calculate Density Altitude :ow to Calculate Density Altitude on a Flight Computer. Learn to Fly at Golden State Flying Club at Gillespie Field in El Cajon, California.