Grade 12 Chemistry
Complete the following equations:
a) 226 88 Ra = 222 86 Rn + ?
b) 234 90 Th = 234 91 Pa + ?
c) 210 84 Po = ? + 4 2 He
d) 24 11 Na = ? + 0 -1 e
e) 239 94 Pu + 4 2 He + ? + 3 1 0 n
a) 226 88 Ra = 222 86 Rn + 4 2 He
b) 234 90 Th = 234 91 Pa + 0 -1 e
c) 210 84 Po = 206 82 Pb + 4 2 He
d) 24 11 Na = 24 10 Ne + 0 -1 e
e) 239 94 Pu + 4 2 He + 1 0 n = 242 96 Cm
a) To balance the equation, we need to ensure that the sum of the atomic numbers and the sum of the mass numbers on both sides of the equation are equal.
226 88 Ra = 222 86 Rn + ?
Since Ra-226 has an atomic number of 88, it means that Rn-222 must have an atomic number of 86 (to balance the equation). Therefore, the missing particle is a helium nucleus (2 2 He).
The balanced equation is:
226 88 Ra = 222 86 Rn + 4 2 He
b) Similar to the previous example, we need to balance the atomic numbers and the mass numbers.
234 90 Th = 234 91 Pa + ?
Since Th-234 has an atomic number of 90, it means that Pa-234 must have an atomic number of 91 (to balance the equation). Therefore, the missing particle is a negative beta particle (0 -1 e).
The balanced equation is:
234 90 Th = 234 91 Pa + 0 -1 e
c) In this equation, we have an alpha particle on the right side (4 2 He) and a polonium nucleus on the left side (210 84 Po).
210 84 Po = ? + 4 2 He
Since an alpha particle has a mass number of 4 and an atomic number of 2, and Polonium has a mass number of 210 and an atomic number of 84, the missing particle must have a mass number of 206 and an atomic number of 82. This corresponds to a lead nucleus (206 82 Pb).
The balanced equation is:
210 84 Po = 206 82 Pb + 4 2 He
d) In this equation, we have sodium on the left side (24 11 Na).
24 11 Na = ? + 0 -1 e
Since sodium has an atomic number of 11, it means that the missing particle must have an atomic number of 10 (to balance the equation). Therefore, the missing particle is a positive positron (0 +1 e).
The balanced equation is:
24 11 Na = 10 1 Ne + 0 -1 e
e) In this equation, we have plutonium on the left side (239 94 Pu), an alpha particle (4 2 He), and three neutrons (3 1 0 n).
239 94 Pu + 4 2 He + ? + 3 1 0 n
Similar to previous examples, we need to balance the atomic numbers and the mass numbers. Plutonium has an atomic number of 94, and the alpha particle has an atomic number of 2. Therefore, the missing particle must have an atomic number of 92 (to balance the equation). Since it does not have a charge or a neutron, the missing particle is a uranium nucleus (235 92 U).
The balanced equation is:
239 94 Pu + 4 2 He + 235 92 U + 3 1 0 n
To complete the given equations, we need to balance the nuclear reactions by ensuring that the sum of the atomic numbers and the sum of the mass numbers on both sides of the equations are equal.
Let's go through each equation step by step:
a) 226 88 Ra = 222 86 Rn + ?
In this equation, the Ra decays to Rn, and an unknown particle is emitted. Since the atomic number decreases by 2 (from 88 to 86), we can deduce that the emitted particle has an atomic number of 2, which corresponds to a helium nucleus, 4 2 He. Therefore, the completed equation is:
226 88 Ra = 222 86 Rn + 4 2 He
b) 234 90 Th = 234 91 Pa + ?
In this equation, the Th decays to Pa, and an unknown particle is emitted. Since the atomic number increases by 1 (from 90 to 91), we can deduce that the emitted particle has an atomic number of 1, which corresponds to a proton, 1 1 H. Therefore, the completed equation is:
234 90 Th = 234 91 Pa + 1 1 H
c) 210 84 Po = ? + 4 2 He
In this equation, an unknown particle decays to Po while emitting a helium nucleus. Since the atomic number increases by 2 (from 84 to 86), we can deduce that the unknown particle has an atomic number of 2, which corresponds to a helium nucleus, 4 2 He. Therefore, the completed equation is:
210 84 Po = 206 82 Pb + 4 2 He
d) 24 11 Na = ? + 0 -1 e
In this equation, an unknown particle decays to Na while emitting a beta particle (electron). Since the atomic number increases by 1 (from 11 to 12), we can deduce that the unknown particle has an atomic number of 1, which corresponds to a proton, 1 1 H. Therefore, the completed equation is:
24 11 Na = 24 12 Mg + 0 -1 e
e) 239 94 Pu + 4 2 He + ? + 3 1 0 n
This equation involves the addition of plutonium (Pu), helium (He), an unknown particle, and a neutron (n). The number of neutrons (mass number minus atomic number) remains the same in a nuclear reaction, so the sum of the mass numbers on both sides of the equation must be equal. To balance this equation, we need to know the mass number and atomic number of the unknown particle.
If you provide the information about the atomic number and mass number of the unknown particle, I can assist you in completing the equation.