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satela [25.4K]
2 years ago
3

titanium-51 decays with a half life of 6 minutes. What fraction of the titanium would remain after one hour?​

Chemistry
1 answer:
vivado [14]2 years ago
3 0

Answer:

\large \boxed{9.76 \times 10^{-4}}

Explanation:

Let A₀ = the original amount of titanium-51.

The amount remaining after one half-life is ½A₀.

After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.

After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀, and so on.

We can write a general formula for the amount remaining:

A = A₀(½)ⁿ

where n is the number of half-lives

n = \dfrac{t}{t_{\frac{1}{2}}}

The fraction remaining is

\text{Fraction} = \dfrac{A}{A_{0}} = \left (\dfrac{1}{2}\right)^{n}

Data:

t_{\frac{1}{2}} = \text{6 min}

t = 1 h

Calculations:

(a) Convert the time to minutes

t = \text{1 h} \times \dfrac{\text{60 min}}{\text{1 h}} = \text{60 min}

(b) Calculate n

n = \dfrac{60}{6} = 10

(c) Calculate the fraction remaining

\dfrac{A}{A_{0}} =  \left (\dfrac{1}{2}\right)^{10} =  \mathbf{9.76 \times 10^{-4}}\\\\\text{The fraction of titanium-51 remaining after 1 h is $\large \boxed{\mathbf{9.76 \times 10^{-4}}}$}

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dimaraw [331]

Answer:

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Explanation:

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x / 12.01067 = 1.70 / 37.99687

cross multiply

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2 years ago
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Explanation :

Using neutralization law,

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M_2 = molar concentration of diluted cognac = 0.0844 M

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4 0
2 years ago
The plant food contains nh4)3po4 what tests would you run to verify the presence of the nh4 ion and the po4 ion
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8 0
2 years ago
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stiks02 [169]

Answer:

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7 0
2 years ago
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Answer:

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Explanation:

Data

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Process

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