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maks197457 [2]
3 years ago
11

Radium has a half-life of 1,620 years. In how many years will a 1 kg sample of radium decay and reduce to 0.125 kg of radium?

Chemistry
2 answers:
JulsSmile [24]3 years ago
8 0

Answer:

4860 years

Explanation:

From

N/No = (1/2)^t/t1/2

Where:

No= mass as time t=0

N= mass at time t

t= time

t1/2= half life

0.125/1 = (1/2)^t/1620

1/8 = (1/2)^t/1620

(1/2)^3= (1/2)^t/1620

3= t/1620

t= 3×1620

t= 4860 years

Tasya [4]3 years ago
7 0
We can calculate years by using the half-life equation. It is expressed as:

A = Ao e^-kt

<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.

</span>From the half-life data, we can calculate for k.

1/2(Ao) = Ao e^-k(1620)
<span>k = 4.28 x 10^-4
</span>
0.125 = 1 e^-<span>4.28 x 10^-4 (</span>t)
t = 4259 years
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Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of
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Explanation:

  • From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525  kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

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4 0
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Given the reaction: 4A1 + 302 - 2A1,O, How many grams of Al will be
dlinn [17]

Answer:

The answer to your question is 27 g of Al

Explanation:

Data

mass of Al = ?

moles of Al₂O₃ = 0.5

The correct formula for the product is Al₂O₃

Balanced chemical reaction

               4Al + 3O₂   ⇒   2Al₂O₃

Process

1.- Calculate the molar mass of the product

Al₂O₃ = (27 x 2) + (16 x 3)

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2.- Convert the moles of Al₂O₃ to grams

               102 g ---------------- 1 mol

                 x       ----------------  0.5 moles

                 x = (0.5 x 102) / 1

                 x = 51 g of Al₂O₃

3.- Use proportions to calculate the mass of Al

                4(27) g of Al --------------- 2(102) g of Al₂O₃

                     x               --------------- 51 g

                        x = (51 x 4(27)) / 2(102)

                        x = 5508 / 204

                        x = 27 g of Al

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