Part C
- D The atomic number is the number of protons in the nucleus, and the
mass number is the sum: number of protons + number of neutrons. If the
atomic number is 39 and the mass number is 89, then the number of
neutrons in the nucleus must be (89) − (39) = 50.
- C Use the plug-in balancing strategy. Since there are at least 4 carbon
atoms on the left, the coefficient of CO 2 cannot be 2, so eliminate (A). If
the coefficient of carbon is 4, we must place a 1 in front of C 4 H 10 to keep
carbons in balance. This will give 10 hydrogens on the left. If we put a 5
in front of H 2 O on the right, we then have 13 oxygens on the right. The
only way we can get 13 oxygens on the left is to place in front of
O 2 on the left. This puts all the elements in balance, but violates the rule
of using only whole numbers, so (B) is wrong. However, if we multiply
the coefficients we just determined by 2, we will maintain balance and
have all whole numbers. So the balanced equation becomes: 2C 4 H 10 (g) +
13O 2 (g) → 8CO 2 (g) + 10H 2 O(l).
- C The mass of a proton is approximately 1 amu, and this is very nearly the
mass of a neutron. Both a positron and an electron are much lighter than 1
amu. A hydrogen molecule weighs roughly twice as much as a proton,
and an alpha particle weighs about four times as much.
- D Add up the mass of 1 mole of this substance. From the periodic table, we
know that
• 1 mole of hydrogen atoms has a mass of about 1 g.
• 1 mole of chlorine atoms has a mass of about 35 g.
• 4 moles of oxygen atoms have a mass of about 64 g.
