Answer:
of sodium arsenate would be present in a 1.70 L sample of drinking water that just meets the standard.
Explanation:
Mass of sodium arsenate =x
Mass of water = M
Density of water = d = 1000 g/L
Volume of the water , V= 1.70 L
M = ![d\times V= 1000 g/L\times 1.70 L=1,700 g=1.700 kg](https://tex.z-dn.net/?f=d%5Ctimes%20V%3D%201000%20g%2FL%5Ctimes%201.70%20L%3D1%2C700%20g%3D1.700%20kg)
1 Parts per billion = 1 μg/kg = ![\frac{10^{-9} kg}{1 kg}](https://tex.z-dn.net/?f=%5Cfrac%7B10%5E%7B-9%7D%20kg%7D%7B1%20kg%7D)
![10 ppb = \frac{x}{1.700 kg}](https://tex.z-dn.net/?f=10%20ppb%20%3D%20%5Cfrac%7Bx%7D%7B1.700%20kg%7D)
![x=10\times 10^{-9} \times 1.700 kg=1.700\times 10^{-8} kg](https://tex.z-dn.net/?f=x%3D10%5Ctimes%2010%5E%7B-9%7D%20%5Ctimes%201.700%20kg%3D1.700%5Ctimes%2010%5E%7B-8%7D%20kg)
![x=1.700\times 10^{-8} kg=1.700\times 10^{-8}\times 10^3 g=1.700\times 10^{-5}g](https://tex.z-dn.net/?f=x%3D1.700%5Ctimes%2010%5E%7B-8%7D%20kg%3D1.700%5Ctimes%2010%5E%7B-8%7D%5Ctimes%2010%5E3%20g%3D1.700%5Ctimes%2010%5E%7B-5%7Dg)
(1 kg =1000 g)
Moles of arsenic =![\frac{1.700\times 10^{-5}g}{75 g/mol}=2.267\times 10^{-7} mol](https://tex.z-dn.net/?f=%5Cfrac%7B1.700%5Ctimes%2010%5E%7B-5%7Dg%7D%7B75%20g%2Fmol%7D%3D2.267%5Ctimes%2010%5E%7B-7%7D%20mol)
1 mole of sodium arsenate has 1 mol of arsenic atom.Then
of arsenic will found in:
of sodium arsenate.
Mass of
of sodium arsenate:
![208 g/mol\times 2.267\times 10^{-7} mo=4.71\times 10^{-5} g](https://tex.z-dn.net/?f=208%20g%2Fmol%5Ctimes%202.267%5Ctimes%2010%5E%7B-7%7D%20mo%3D4.71%5Ctimes%2010%5E%7B-5%7D%20g)
of sodium arsenate would be present in a 1.70 L sample of drinking water that just meets the standard.
Answer:
![0.28 \mu g](https://tex.z-dn.net/?f=0.28%20%5Cmu%20g)
Explanation:
We can write the amount of mass of radon-222 left after a time t by using the equation:
![m(t) = m_0 (\frac{1}{2})^{t/t_{1/2}}](https://tex.z-dn.net/?f=m%28t%29%20%3D%20m_0%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7Bt%2Ft_%7B1%2F2%7D%7D)
where
is the initial mass
t is the time
is the half-life
Substituting t = 15.2 d in the formula, we find
![m(15.2 d) = (4.38 \mu g) (\frac{1}{2})^{15.2d/3.82 d}=0.28 \mu g](https://tex.z-dn.net/?f=m%2815.2%20d%29%20%3D%20%284.38%20%5Cmu%20g%29%20%28%5Cfrac%7B1%7D%7B2%7D%29%5E%7B15.2d%2F3.82%20d%7D%3D0.28%20%5Cmu%20g)
All are the same elements but they differ in their atomic mass therefore this process will called Isotopes
B. Rovers are what you are looking for