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Shtirlitz [24]
2 years ago
15

One of the nuclides in spent nuclear fuel is U-235 , an alpha emitter with a half-life of 703 million years. How long will it ta

ke for an amount of U-235 to reach 29.0% of its initial amount
Chemistry
1 answer:
sergey [27]2 years ago
8 0

It will take 1.254 billion years to reach 29% of the original amount of U-235.

<h3>First-order Radioactive Decay:</h3>

The rate constant can be calculated from the half-life. The relationship of the two is shown below:

t_{0.5} =\frac{In 2}{k}

k=\frac{In 2}{703}

 = 9.86 × 10 ^-^{4} my^{-1}

The first-order integrated rate law of a nuclide undergoing radioactive decay is;

ln[A]_t = -kt + ln[A]_0.

ln 0.29 = ln1 - 9.86 × 10^-^4 t

ln0.29= -9.86 × 10^-^4

t= 1254.5 my

It will take 1.254 billion years to reach to 29% of the original amount.

Learn more about radioactive decay here:

brainly.com/question/9932896

#SPJ4

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Initial dimension of the pool:

Length (L)= 50.0 m

Width (W)= 25.0 m

Depth (D) = 2.0 m

Initial Volume of the pool (V1)  = L*W*D = 50.0 * 25.0 *2.0 = 2500 m3

When the depth is lowered by 3 cm i.e. 0.03 m, the new depth becomes: 2.0 - 0.03 = 1.97 m

The new volume of the pool (V2) = 50.0*25.0*1.97 = 2462.5 m3

Volume of water to be pumped out = V1-V2 = 2500-2462.5 = 35.5 m3

Now, 1 m3 = 1000 L

therefore, 35.5 m3 == 35.5 *10^3 L

It is given that:

3.80 L of water is pumped out in 1 sec

Therefore, 35.5*10^3 L will be pumped out in:

= 1 s * 35.5*10^3 L/3.80 L = 9.34*10^3 sec



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3 years ago
The solid in question 2 was aluminum sulfate, al2(so4)3. calculate the molar heat of solution, δhsoln, for aluminum sulfate. hin
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The solution for this problem is:

The information given lacks but nevertheless the answer is:
So, the total heat free by dissolving the solute was 1386 + 32 = 1417 J
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Which element is likely more reactive, and why
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Combustion of hydrocarbons such as butane (C4H10) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's a
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Explanation

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According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

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