"\\bold{Given:}\\\\\nHg\\\\\nV = 5 cm^3\\\\\n\\rho = 13.55 g\/cm^3\\\\[1ex]\n\n\\bold{Find:}\\\\\nN = \\ ??\\\\[3ex]\n\nN = \\nu N_A = \\dfrac{mN_A}{M} = \\dfrac{\\rho VN_A}{M} = \\dfrac{13.55g\/cm^3*5cm^3*6.02*10^{23}mol^{-1}}{200.59 g\/mol} \\approx 2.0333*10^{23}\\\\[1ex]\n\\bold{Answer:} \\ 2.0333*10^{23} \\ atoms"
Comments
Leave a comment