Chemistry, 24.12.2020 23:00 Yamari000

A 500 mL sample of drinking water contains 0.001 mg of mercury. How many ppb is this?

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Answer from: polyanskiymichael

The value is $k = 2 \ ppb$

Explanation:

From the question we are told that

The volume of the drinking water is $V = 500 \ mL = 500 *10^{-3} \ L = 500 *10^{-6} \ m^3$

The mass of mercury it contains $m_c = 0.001 \ mg = 0.001 *10^{-3} \ g = 0.001 *10^{-6}\ kg$

Generally the mass of water is mathematically represented as

$m = V * \rho$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg/m^3$

$m = 500 *10^{-6 }* 1000$

=> $m = 0.5 \ kg$

Generally ppb mean part per billon , and 1 billon is 100,000,000

So the part per billon which this represents is mathematically represented as

$k = \frac{0.001 *10^{-6}}{ 0.5} * 100 000 000$

=> $k = 2 \ ppb$

Answer from: Quest

Idon't know hope someone you

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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