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andrezito [222]
2 years ago
13

The element rhenium (Re) has two naturally occurring isotopes 185Re and 187Re. Rhenium is 62.60% 187 Re (186.956 amu) and 185Re

(184.955 amu). Calculate the average atomic mass.
Chemistry
1 answer:
Gre4nikov [31]2 years ago
5 0

The average atomic mass of the element rhenium (Re) is 186.208 amu

<h3>Data obtained from the question</h3>

Ffrom the question given above, the following data were obtained:

  • Mass of isotope A = 184.955 amu
  • Mass of isotope B = 186.956 amu
  • Abundance of B = 62.60%
  • Abundance of A = 100 - 62.60 = 37.4%
  • Average atomic mass =?

<h3>How to determine the average atomic mass </h3>

The average atomic mass of the element can be obtained as illustrated below:

Average atomic mass = [(Mass of A × Abundance of A)/100] + [(Mass of B × Abundance of B)/100]

Average atomic mass = [(184.955 × 37.4)/100] + [(186.956 × 62.60)/100]

Average atomic mass = 186.208 amu

Thus, the average atomic mass of the element is 186.208 amu

Learn more about isotope:

brainly.com/question/28223231

#SPJ1

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The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

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<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

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