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gavmur [86]
3 years ago
6

HELP PLEASE, I NEED THIS BY TODAY!

Chemistry
1 answer:
labwork [276]3 years ago
6 0

Answer:

2.48 g

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 10 g

Time (t) = 1407.6 million years

Amount remaining (N) =?

Next, we shall determine the rate of decay (K) of uranium-235. This can be obtained as follow:

NOTE: Uranium-235 has a half life of 700 million years.

Decay constant (K) =?

Half life (t½) = 700 million years

K = 0.693/t½

K = 0.693/700

K = 9.9×10¯⁴ / year

Therefore, Uranium-235 decay at a rate of 9.9×10¯⁴ / year.

Finally, we shall determine the amount of Uranium-235 remaining after 1407.6 million years as follow:

Original amount (N₀) = 10 g

Time (t) = 1407.6 million years

Decay constant (K) = 9.9×10¯⁴ / year

Amount remaining (N) =?

Log (N₀/N) = kt /2.3

Log (10/N) = (9.9×10¯⁴ × 1407.6) /2.3

Log (10/N) = 0.60588

10/N = antilog (0.60588)

10/N = 4.04

Cross multiply

10 = 4.04 × N

Divide both side by 4.04

N = 10/4.04

N = 2.48 g

Therefore, 2.48 g of uranium-235 is remaining after 1407.6 million years.

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Answer:

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<u>Step 1: </u>Data given

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Let's suppose  x = mass of CaO. Then mass of BaO = 4.10-x.

In the balanced equation, we notice the mole ratio of CaO/BaO with CO2 is 1:1. This means the number of moles CaO + moles BaO = Number of moles CO2

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