Key for CR #7 1. HCl(aq) + KOH(aq) KCl + H2O 2. (1.22 M)(10.00 mL) / 15.65 mL = the answer 3. Alkyl halide or halocarbon 4. Substitution 5. It has only single carbon-carbon bonds. 6. 2-iodo-2-methylpropane is more polar than 2-methylpropane. Therefore 2-iodo-2methylpropane has stronger intermolecular forces than 2-methylpropane. Therefore 2-iodo-2methylpropane has a higher boiling point. 7. Na 8. 8 moles of C2H6 (7 moles O2 / 2 moles C2H6) = 28 moles O2 9. 2HI H2 + I2 10. Concentrations have reached constant levels. Concentrations are not changing. 11. (0.013 mol I2 /L) x 1.00 L x (254 g I2 / mole I2) = 3.3 g I2. Note: the concentration of the I2 and the volume of the flask are given in the reading passage. The student must calculate the molar mass of I2. 12. 4 2He + 237 93Np (see Table N for decay mode of Am-241) 13. Am-241 has a longer half-life than Fr-220 so the sample emits radioactive particles for a long time. (That is, the smoke detector will last longer.) Fr-220 has a short half life and decays quickly. (Info about half-lives in table N.) 14. Smoke particles interrupt the flow of ions required to maintain an electric current. (See final sentence of reading passage.)