## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$q = -5.36\times 10^7~C$
We can find the mass of the water. $m = \rho~V$ $m = (1000~kg/m^3)(1.0~L)(\frac{1~m^3}{10^3~L})$ $m = 1.0~kg$ Each water molecule has 10 protons and 8 neutrons. We can find the number $N$ of water molecules. $N = \frac{1.0~kg}{(18)(1.66\times 10^{-27}~kg)}$ $N = 3.35\times 10^{25}$ Each water molecule has ten electrons. Each electron has a charge of $-1.6\times 10^{-19}~C$. We can find the total charge of all the electrons. $q = (3.35\times 10^{25})(10)(-1.6\times 10^{-19}~C)$ $q = -5.36\times 10^7~C$