Answer:
<em><u>Question 1.</u></em><em> → </em>The amount remained is 15.625 g after 5 half-lives (30 hours).
<u><em>Question 2. </em></u><em>→ </em>mass % of water =41.4 %
<u><em>Question 3. </em></u><em>→ </em>C. 64 grams.
Explanation:
<em><u>Question 1. </u></em>
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).
<u><em>Question 2.</em></u>
- 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 %
<em><u>Question 3. </u></em>
<u>Mass percentage</u> 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:
b) 47ml
Explanation:
The beaker would look like this:
60 mL
50 mL
------- The bottom of the meniscus is here
40 mL
30 mL
So <u>we have between 40 and 50 mL</u>. There are 10 mL of "distance" between those two values, and three fourths of that difference is (10 * 3/4) 7.5 mL.
So <em>the volume is 47.5 mL</em>, but because the measurement instrument (the beaker) does not handle decimals, the value we should report is 47 mL.