Key for CR #7

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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.)
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