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bogdanovich [222]
2 years ago
6

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

are the average percent abundances of each isotope?
Chemistry
1 answer:
Lina20 [59]2 years ago
6 0

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

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For the reaction NH4NO3 (s) → N2O (g) + 2H2O (l), you decompose 1 mole NH4NO3 and only get 0.75 moles of N2O. The percent yield
attashe74 [19]

Answer:

The answer to your question is False.

Explanation:

Data

1 mole of NH₄NO₃

0.75 moles of N₂O

Percent yield = 25%

Chemical reaction

               NH₄NO₃   ⇒   N₂O  +  2H₂O

Process

1.- Determine the theoretical yield

              1 mol NH₄NO₃ ------------- 1 mol of N₂O

2.- Calculate the percent yield

             Percent yield = Actual yield / Theoretical yield  x 100

-Substitution

              Percent yield = 0.75 / 1 x 100

-Simplification

              Percent yield = 0.75 x 100

-Result

              Percent yield = 75%

Conclusion

False, the actual percent yield is 75%

5 0
3 years ago
Suppose you were to make a ring from a single strand of gold atoms. How many gold atoms would be required to make such a ring? H
Rudiy27
Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:

54.3 mm = 5.43 cm

Now, we determine the number of gold atoms that will be present in this:

5.43 / 1 x 10⁻⁹

There will be 5.43 x 10⁹ atoms


We now determine the number of moles this is by:

one mole = 6.02 x 10²³ atoms

Moles = 5.43 x 10⁹ / 6.02 x 10²³ 
Moles = 9.01 x 10⁻¹⁵ moles

The molar mass of gold is 197 g/mol

The mass is 9.01 x 10⁻¹⁵  * 197

The mass of the strand is 1.76 x 10⁻¹² grams
8 0
3 years ago
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7 0
2 years ago
How many H2O molecules are in 183.2 grams of H20 gas?
jek_recluse [69]

Answer: There are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of H_{2}O are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol

According to the mole concept, there are 6.022 \times 10^{22} molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}

Thus, we can conclude that there are 61.24 \times 10^{23} molecules present in 183.2 grams of H_{2}O gas.

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How much concentrated 5.0M nitric acid is needed to prepare 50 mL of a 1.0 M solution?
Ksivusya [100]

Explanation:

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