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grigory [225]
3 years ago
11

The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later

Chemistry
1 answer:
Kipish [7]3 years ago
3 0

Answer:

12.02 g

Explanation:

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

Half life (t½) = 2 days

Original amount (N₀) = 96 g

Time (t) = 6 days

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 2

K = 0.3465 /day

Therefore, the rate of disintegration of the isotope is 0.3465 /day.

Finally, we shall determine the amount of the isotope remaining after 6 days as follow:

Original amount (N₀) = 96 g

Time (t) = 6 days

Decay constant (K) = 0.3465 /day.

Amount remaining (N) =.?

Log (N₀/N) = kt / 2.303

Log (96/N) = (0.3465 × 6) / 2.303

Log (96/N) = 2.079/2.303

Log (96/N) = 0.9027

Take the anti log of 0.9027

96/N = anti log (0.9027)

96/N = 7.99

Cross multiply

96 = N × 7.99

Divide both side by 7.99

N = 96 /7.99

N = 12.02 g

Therefore, the amount of the isotope remaining after 6 days is 12.02 g

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3 years ago
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
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Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

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[AgNO3] = 0.20 M

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Ksp of BaCrO4 = 1.1 x 10^-10

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Ksp=[Ba^{2+}][CrO_{4}^{2-}]

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Now,

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Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

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