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Alik
28 days ago
13

naturally occurring bromine molecules, br2 have masses of 158, 160, and 162. they occur in the relative abundances 25.69%, 49.99

%, and 24.31% respectively. what is the average atomic mass of bromine atoms? what is the relative abundance of 79br and 81br isotopes?
Chemistry
1 answer:
Tems11 [2.4K]28 days ago
4 0
To calculate the average atomic mass of an element, you can multiply the masses of its isotopes by their corresponding relative abundances and sum them up.

Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu

According to the information given, the relative abundance of Br-79 stands at 25.69% because two atoms of Br yield 79*2 = 158 amu. Likewise, for Br-81, the relative abundance is 162 due to 81*2 = 162, which is 24.31%.
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The forward reaction will keep occurring until all NO or all NO₂ is consumed.

Clarification:

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  • Thus, removing the product (N₂O₃) from the system effectively lowers the product concentration, prompting the reaction to shift forward and generate additional product in order to alleviate the strain caused by the removal of N₂O₃.

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To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if
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The correct options include choice 2, 3, and 6.

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Density is identified as the mass of a substance per unit volume occupied by that substance.

Density=\frac{Mass}{Volume}

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2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.

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The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in  15.3 \frac{atoms}{min-g}

The dissolution rate of the sample is given by 9.16 \frac{atoms} {min-gram}

The C-14 proportion within the sample can be determined as per = \frac {9.16}{15.3} \\\\ = 0.5987

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Now, we need to compute the number of half-lives (n) that are applicable:

(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\

\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740\\

Thus, the age of the sample is represented as = n \times\ half-life

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