Question 1.
an isotope has a half-life of 6 hours. if the original sample had a mass of 500 grams, how much of the sample would be left after 30 hours. be sure to identify the number of 1/2 life that occurred and to explain how you obtained this answer.

question 2.
calculate the percent by mass of water in copper (ii) fluoride tetra hydrate. you must show all of your work to earn credit. atomic masses of each element: cu = 64, f = 19, o = 16, h= 1

question 3.

if the percent (mass/mass) for a solute is 16% and the mass of the solution is 400 g, what is the mass of solute in solution?

a.6400 grams
b. 25 grams
c.64 grams
d. 0.04 grams

Question 1.  → The amount remained is 15.625 g after 5 half-lives (30 hours).

Question 2. →  mass % of water =41.4 %

Question 3. →  C. 64 grams.

Explanation:

Question 1.

It is known that the decay of isotopes and radioactive material obeys first order kinetics.

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

Thus, 30 hours represent (30/6) 5 half-lives.

500 g →(first 1/2 life) 250 g →(second 1/2 life) 125 g →(third 1/2 life) 62.5 g →(fourth 1/2 life) 31.25 g →(fifth 1/2 life) 15.625 g

So, The amount remained is 15.625 g after 5 half-lives (30 hours).

Question 2.

mass % of water = (mass of 4 mol of H₂O) / (mass of 1 mol of CuF₂.4H₂O ) *100 = (72 /174) *100 = 41.4 %

Question 3.

Mass percentage is defined as the mass of a solute divided by the total mass of the solution, multiplied by 100

∴ 16 = ( mass of solute / 400 g) *100

mass of solute = (16* 400)/ 100 = 64 g.

So the right choice is C. 64 grams.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

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