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Liula [17]
3 years ago
10

The table shows the amount of radioactive element remaining in a sample over a period of time.

Chemistry
1 answer:
Lunna [17]3 years ago
6 0

Answer:

1. The half-life is 22 years.

2. 132 years

Explanation:

1. Determination of the the half-life.

The half-life of an element is the time taken for half the element to decay.

From the table given above, the original amount of the element 45 g. If we divide 45 by 2, we'll have 22.5 g as half the original amount of element.

Now, the time taken to obtain 22.5 g as shown from the table is 22 years.

Thus, the half-life the element is 22 years.

2. Determination of the time.

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Time (t) =?

Next, we shall the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 308 g

Amount remaining (N) = 4.8125 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

4.8125 = 1/2ⁿ × 308

Cross multiply

4.8125 × 2ⁿ = 308

Divide both side by 4.8125

2ⁿ = 308 / 4.8125

2ⁿ = 64

Express 64 in index form with 2 as the base.

2ⁿ = 2⁶

n = 6

Thus, 6 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Number of half-lives (n) = 6

Half-life (t½) = 22 years

Time (t) =?

n = t / t½

6= t / 22 years

Cross multiply

t = 6 ×22

t = 132 years.

Thus, the time taken is 132 years.

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3 years ago
If 4520 kj of heat is needed to boil a sample of water, what is the mass of water
Cerrena [4.2K]

Answer:

1,085g of water

Explanation:

If we have the value 4520kj is because the question is related to Energy and heat capacity. In this case, the law and equation that we use is the following:

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In order to determine the mass, I will have the same equation but finding m

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Answer:

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Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

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