Chemistry, 03.02.2020 22:02 btsheygirl

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

Answers: 3

Show answers

Answers

Answer from: rorathan123

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.

The formula of Copper (II) Fluoride tetra hydrate is CuF₂.4H₂O.the molar mass of Copper (II) Fluoride tetra hydrate is the sum of the atomic masses o each element.molar mass of 1 mol of CuF₂.4H₂O = 64 + 2*(19) + 8* (1) + 4*(16) = 174 g/mol.molar mass of 4 mol of H₂O = 8*(1) + 4*(16) = 72 g/mol 1 mol of Copper (II) Fluoride tetra hydrate → 4 mol H₂O

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

mass percent = ( mass of solute / mass of solution) *100

∴ 16 = ( mass of solute / 400 g) *100

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

So the right choice is C. 64 grams.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

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