Select all the discoveries for which ernest rutherford is credited. the emissions of radioactive elements the neutron the electr
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Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
Hence, the percent abundance of O-18 is:
Hence, the percent abundance of oxygen-18 is:
Therefore, the percent abundance of oxygen-18 is 1.9066%.
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Taking into account the reaction stoichiometry and ideal gas law, 0.48144 grams of KClO₃ was decomposed.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 KClO₃ → 2 KCl + 3 O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- KClO₃: 2 moles
- KCl: 2 moles
- O₂: 3 moles
The molar mass of the compounds is:
- KClO₃: 122.45 g/mole
- KCl: 74.45 g/mole
- O₂: 32 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- KClO₃: 2 moles ×122.45 g/mole= 244.8 grams
- KCl: 2 moles ×74.45 g/mole= 148.9 grams
- O₂: 3 moles ×32 g/mole= 96 grams
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
- P is the gas pressure.
- V is the volume that the gas occupies.
- T is the temperature of the gas.
- R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
- n is the number of moles of the gas.
165 mL of oxygen is produced at 30.0 °C and 90.0 kPa. This is, you know:
- P= 90 kPa= 0.888231 atm (being 101.325 kPa= 1 atm)
- V= 165 mL= 0.165 L (being 1000 mL= 1 L)
- n= ?
- R= 0.082
- T= 30 C= 303 K (being 0 C= 273 K)
Replacing in the ideal gas law:
0.888231 atm× 0.165 L = n× 0.082 × 303 K
Solving:
n= (0.888231 atm× 0.165 L)÷ (0.082 × 303 K)
<u><em>n= 0.0059 moles</em></u>
Finally, 0.0059 moles of oxygen is produced at 30 °C and 90 kPa.
<h3>Mass of KClO₃ required</h3>The following rule of three can be applied: If by stoichiometry of the reaction 3 moles of O₂ are produced by 244.8 grams of KClO₃, 0.0059 moles of O₂ are produced by how much mass of KClO₃?
<u><em>mass of KClO₃= 0.48144 grams</em></u>
Finally, 0.48144 grams of KClO₃ was decomposed.
Learn more about
the reaction stoichiometry:
<u>brainly.com/question/24741074</u>
<u>brainly.com/question/24653699</u>
ideal gas law: