Answer:
₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴
Explanation:
The given nuclear reaction shows alpha decay.
₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴
Properties of alpha radiations:
Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.
Alpha radiations can travel in a short distance.
These radiations can not penetrate into the skin or clothes.
These radiations can be harmful for the human if these are inhaled.
These radiations can be stopped by a piece of paper.
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy
Answer:
24.32
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 24
Abundance (A%) = 78.70%
Isotope B
Mass of B = 25
Abundance (B%) = 10.13%
Isotope C:
Mass of C = 26
Abundance (C%) = 11.17%
Average atomic mass of Mg =..?
The average atomic mass of Mg can be obtained as illustrated below:
Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]
= 18.888 + 2.5325 + 2.9042
= 24.3247 ≈ 24.32
Therefore, the average atomic mass of magnesium (Mg) is 24.32
Answer:
63.05% of MgCO3.3H2O by mass
Explanation:
<em>of MgCO3.3H2O in the mixture?</em>
The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:
<em>Mass water:</em>
3.883g - 2.927g = 0.956g water
<em>Moles water -18.01g/mol-</em>
0.956g water * (1mol/18.01g) = 0.05308 moles H2O.
<em>Moles MgCO3.3H2O:</em>
0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =
0.01769 moles MgCO3.3H2O
<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>
0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O
<em>Mass percent:</em>
2.448g MgCO3.3H2O / 3.883g Mixture * 100 =
<h3>63.05% of MgCO3.3H2O by mass</h3>
The Volumes can be calculated from Masses by using following Formula,
Density = Mass / Volume
Solving for Volume,
Volume = Mass / Density
Mass of Both Gases = 14.1 g
Density of Argon at S.T.P = 1.784 g/L
Density of Helium at S.T.P = 0.179 g/L
For Argon:
Volume = 14.1 g / 1.784 g/L
Volume = 7.90 L
For Helium:
Volume = 14.1 g / 0.179 g/L
Volume = 78.77 L
