# Homework 9

- Page ID
- 28866

**Q1.65 **

__Question__ Give the number of protons and neutrons in each isotope listed

\[_{19}^{41}\textrm{K}\]

\[_{4}^{11}\textrm{Be}\]

\[_{22}^{45}\textrm{Ti}\]

\[_{8}^{16}\textrm{O}\]

__Solution__

**Strategy:**

Take the larger number and subtract the smaller number from it. The smaller number is the atomic number (z), which indicates the number of protons the element has. The larger number indicates the number of protons and neutron present in the element.

**a. **^{41}K has an atomic mass of 41. The atomic mass is the number of protons and neutrons present in the element. To get the number of neutrons, subtract the atomic number (19) from the atomic mass (41). When done so you find that K^{41} has 19 protons and 22 neutrons. Keep in mind the atomic number is the number of protons present in the element, so the only thing you are trying to find through subtraction is the number of neutrons.

**b. **For ^{11}Be, the same thing occurs. Take the atomic number and subtract it from the atomic mass. The atomic mass is 11 and the atomic number is 4, so, 11 minus for equals 7. The answer for part **b.** is 4 protons and 7 neutrons.

**c. **For ^{45 }Ti, you subtract 22 from 45, which gives you 22 protons and 23 neutrons.

**d. **For ^{16}O, the same method is applied. Subtract 8 from 16 and you get 8 protons and 8 electrons.

**Q2.20**

__Question __A silver picture frame displaces 0.454 L of water and has a mass of 2.78 kg. Calculate the density of the silver in g/cm^{3}.

__Solution__

**Strategy:**

**a. **Convert L to cm^{3} using dimensional analysis and conversion factors.

\[\frac{0.454L}{1} X \frac{1,000 mL }{1 L} X \frac{1 cm^{3}}{1 mL} = 454cm^{^{3}}\]

**b. **Convert kg to g using dimensional analysis and conversion factors.

\[\frac{2.78kg}{1} X \frac{1,000 g }{1 kg} X = 2,780g^{^{}}\]

**c. **Calculate the density.

\[D=\frac{m}{v}\]

\[=\frac{2,780g}{454cm^{^{3}}}\]

\[=6.12\frac{g}{cm^{^{3}}}\]