Complete each nuclear reaction. (}_{1)^2H+_(1)^2Harrow _(E)^0He+_(0)^1n (}_{92)^238Uarrow _(6)^6Th+_(2)^4He D: square F: square E: square G square

Let's correct the reactions and balance them properly:<br /><br />1. For the first reaction:<br />\[ {}_{1}^{2}H + {}_{1}^{2}H \rightarrow {}_{2}^{4}He + 2 \, {}_{0}^{1}n \]<br /><br />So, the correct reaction is:<br />\[ {}_{1}^{2}H + {}_{1}^{2}H \rightarrow {}_{2}^{4}He + 2 \, {}_{0}^{1}n \]<br /><br />2. For the second reaction:<br />\[ {}_{92}^{238}U \rightarrow {}_{90}^{234}Th + {}_{2}^{4}He \]<br /><br />So, the correct reaction is:<br />\[ {}_{92}^{238}U \rightarrow {}_{90}^{234}Th + {}_{2}^{4}He \]<br /><br />Now, let's identify the variables:<br /><br />- D: The product nucleus in the second reaction is \({}_{90}^{234}Th\).<br />- F: The emitted particle in the second reaction is \({}_{2}^{4}He\).<br />- E: The product nucleus in the first reaction is \({}_{2}^{4}He\).<br />- G: The emitted particle in the first reaction is \({}_{0}^{1}n\).<br /><br />So, the completed reactions are:<br /><br />1. \[ {}_{1}^{2}H + {}_{1}^{2}H \rightarrow {}_{2}^{4}He + 2 \, {}_{0}^{1}n \]<br />2. \[ {}_{92}^{238}U \rightarrow {}_{90}^{234}Th + {}_{2}^{4}He \]<br /><br />And the variables are:<br />- D: \({}_{90}^{234}Th\)<br />- F: \({}_{2}^{4}He\)<br />- E: \({}_{2}^{4}He\)<br />- G: \({}_{0}^{1}n\)