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Artist 52 [7]
3 years ago
9

Does the size of the sample of uranium-238 affect its half-life

Chemistry
1 answer:
AleksandrR [38]3 years ago
3 0
Answer is: <span>the size of the sample of uranium-238 does not affect its half-life.
</span><span>
The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration (size of the sample).
</span>Half-life <span>is the time required for a quantity (in this example number of radioactive uranium) to reduce to half its initial value.</span><span>

</span>
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3 years ago
If 0.2 g of nitrobenzene are added to 10.9 g of naphthalene, calculate the molality of the solution. (given: molar mass of nitro
In-s [12.5K]

Molality is defined as 1 mole of a solute in 1 kg of solvent.  

Molality=

\frac{Number of moles of solute}{Mass of solvent in kg}

Number of moles of solute, n=  

\frac{Given mass of the substance}{Molar mass of the substance}

Given mass of the nitrobenzene=0.2 g

Molar mass of the substance= 123.06 g mol⁻¹

Number of moles of nitrobenzene,  

n= \frac{0.2 }{123.06}

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7 0
3 years ago
Silver (Ag) has two stable isotopes: 107Ag, 106.90 amu, and 109Ag, 108.90 amu. If the average atomic mass of silver is 107.87 am
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Answer:

  • The abundance of 107Ag is 51.5%.
  • The abundance of 109Ag is 48.5%.

Explanation:

The <em>average atomic mass</em> of silver can be expressed as:

107.87 = 106.90 * A1 + 108.90 * A2

Where A1 is the abundance of 107Ag and A2 of 109Ag.

Assuming those two isotopes are the only one stables, we can use the equation:

A1 + A2 = 1.0

So now we have a system of two equations with two unknowns, and what's left is algebra.

First we<u> use the second equation to express A1 in terms of A2</u>:

A1 = 1.0 - A2

We <u>replace A1 in the first equation</u>:

107.87 = 106.90 * A1 + 108.90 * A2

107.87 = 106.90 * (1.0-A2) + 108.90 * A2

107.87 = 106.90 - 106.90*A2 + 108.90*A2

107.87 = 106.90 + 2*A2

2*A2 = 0.97

A2 = 0.485

So the abundance of 109Ag is (0.485*100%) 48.5%.

We <u>use the value of A2 to calculate A1 in the second equation</u>:

A1 + A2 = 1.0

A1 + 0.485 = 1.0

A1 = 0.515

So the abundance of 107Ag is 51.5%.

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