One mole of helium atoms has a mass of 4 grams. if a helium atom in a balloon has a kinetic energy of 1.224e-21 j, what is the speed of the helium atom? (the speed is much lower than the speed of light.) v = 6.12e-19 incorrect: your answer is incorrect. m/s

607 m/s First, let's calculate the mass of a helium atom. You do that by dividing the mass of 1 mole of helium atoms by avogadro's number. So 4 / 6.0221409e23 = 6.6422e-24 g = 6.6422e-27 kg The equation for kinetic energy is E = 0.5 M V^2 And we have 1.224e-21 J, which is 1.224e-21 kg*m^2/s^2 So substitute the known values into the equation for kinetic energy and solve for V 1.224e-21 kg*m^2/s^2 = 0.5 6.6422e-27 kg V^2 1.224e-21 kg*m^2/s^2 = 3.32108e-27 kg V^2 368555 m^2/s^2 = V^2 607 m/s = V