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Fofino [41]
4 years ago
8

If the concentration of a saturated solution at 0∘C is 12.5 gCuSO4/100 g soln, what mass of CuSO4⋅5H2O would be obtained? [Hint:

Note that the solution composition is stated in terms of CuSO4 and that the solid that crystallizes is the hydrate CuSO4⋅5H2O.]
Chemistry
1 answer:
Andrew [12]4 years ago
7 0

Answer:

Mass of CuSO4.H2O obtained: m_{total}=19.52 g

Explanation:

The molecular weight of the salt is: M=159.5 g/mol

<u>In the solution</u>: 12.5 g of CuSO4

In moles: n=\frac{12.5 g}{159.5 g/mol}=0.078 mol

<u>Mass of wate</u>r:

5 moles of water per mol of salt: m=\frac{5mol}{1mol}*0.078mol*\frac{18g}{mol}=7.02 g

Mass of CuSO4.H2O obtained: m_{total}=12.5 g + 7.02 g=19.52 g

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Which of the following is is a chemical property of pure water
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The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

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3 years ago
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