Write balanced nuclear equations for (a) the loss of an alpha particle by Th-230. (b) the loss of a beta particle by lead-210. (c) the fission of U-235 to give Ba-140, another nucleus, and an excess of two neutrons. (d) the K-capture of Ar-37.

To write the balanced nuclear equation for the loss of an alpha particle by Th-230, we must make sure that the mass numbers (A) and atomic numbers (Z) on both sides of the equation are equal. Thorium-230 (Th-230) loses an alpha particle, which is a helium-4 nucleus, so we write: Th-230 -> He-4 + X Now we just need to find the nucleus X. An alpha particle consists of 2 protons and 2 neutrons, therefore, the mass number (A) decreases by 4, and the atomic number (Z) decreases by 2. The balanced nuclear equation for the alpha decay of Th-230 is: ^{230}_{90}Th -> ^{4}_{2}He + ^{226}_{88}Ra

To write the balanced nuclear equation for the loss of a beta particle by lead-210 (Pb-210), we must ensure that the mass numbers (A) and atomic numbers (Z) on both sides of the equation are equal. Pb-210 loses a beta particle (electron). A beta particle has a mass number of 0 and an atomic number of -1: Pb-210 -> e-1 + X Now we just need to find the nucleus X. Since a beta particle has a mass number of 0, the mass number (A) remains unchanged. However, the atomic number (Z) increases by 1. The balanced nuclear equation for the beta decay of Pb-210 is: ^{210}_{82}Pb -> ^{0}_{-1}e + ^{210}_{83}Bi

To write the balanced nuclear equation for the fission of uranium-235 (U-235) to give barium-140 (Ba-140), another nucleus, and an excess of two neutrons, we must ensure that the mass numbers (A) and atomic numbers (Z) on both sides of the equation are equal: U-235 -> Ba-140 + Y + 2n Now we need to find the nucleus Y. Since the mass numbers (A) and atomic numbers (Z) must be equal on both sides of the equation, we can set up the following equations for mass number and atomic number: A: 235 = 140 + A_Y + 2(1) Z: 92 = 56 + Z_Y + 2(0) Solving for A_Y and Z_Y, we get: A_Y = 235 - 140 - 2 = 93 Z_Y = 92 - 56 = 36 So, the nucleus Y is ^{93}_{36}Kr. The balanced nuclear equation for the fission of U-235 to give Ba-140, another nucleus, and an excess of two neutrons is: ^{235}_{92}U -> ^{140}_{56}Ba + ^{93}_{36}Kr + 2^{1}_{0}n

To write the balanced nuclear equation for the K-capture of argon-37 (Ar-37), we must take into account that an electron from the inner shell is captured by the nucleus, and as a result, the atomic number (Z) decreases by 1, while the mass number (A) remains unchanged: Ar-37 + e -> X Now we just need to find the nucleus X. The mass number (A) remains the same; however, the atomic number (Z) decreases by 1. The balanced nuclear equation for the K-capture of Ar-37 is: ^{37}_{18}Ar + ^{0}_{-1}e -> ^{37}_{17}Cl