Question

Carry out each calculation, paying special attention to significant figures, rounding, and units \((J=\) joule, the \(S I\) unit of encrgy; mol = mole, the SI unit for amount of substance): (a) \(\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.9979 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{489 \times 10^{-9} \mathrm{~m}}\) (b) \(\frac{\left(6.022 \times 10^{23} \text { molecules } / \mathrm{mol}\right)\left(1.23 \times 10^{2} \mathrm{~g}\right)}{46.07 \mathrm{~g} / \mathrm{mol}}\) (c) \(\left(6.022 \times 10^{23}\right.\) atoms \(\left./ \mathrm{mol}\right)\left(1.28 \times 10^{-18} \mathrm{~J} /\right.\) atom \()\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)\) where the numbers 2 and 3 in the last term are exact

Short answer

a) 4.06 \( \times \) 10^{-19} J, b) 1.61 \( \times \) 10^{24} molecules, c) 1.07 \( \times \) 10^{5} J/mol.

Step-by-step solution

- Calculate Part (a)

Identify the given values and plug them into the equation: \[ \frac{(6.626 \times 10^{-34} \, \mathrm{J} \cdot \mathrm{s})(2.9979 \times 10^{8} \, \mathrm{m} \, / \, \mathrm{s})}{489 \, \times \, 10^{-9} \, \mathrm{m}} \] Calculate the numerator first: \[ 6.626 \, \times \, 10^{-34} \, \mathrm{J} \cdot \mathrm{s} \times 2.9979 \, \times \, 10^{8} \, \mathrm{m} / \mathrm{s} = 1.9863 \, \times \, 10^{-25} \, \mathrm{J} \cdot \mathrm{m} \] Now, divide by the denominator: \[ \frac{1.9863 \, \times \, 10^{-25} \, \mathrm{J} \cdot \mathrm{m}}{489 \, \times \, 10^{-9} \, \mathrm{m}} = 4.06 \, \times \, 10^{-19} \, \mathrm{J} \] The result is \(4.06 \, \times \, 10^{-19} \, \mathrm{J}\).

- Calculate Part (b)

Identify the given values and plug them into the equation: \[ \frac{(6.022 \, \times \, 10^{23} \, \text{molecules} / \, \mathrm{mol})(1.23 \, \times \, 10^{2} \, \mathrm{g})}{46.07 \, \mathrm{g} / \, \mathrm{mol}} \] Calculate the numerator first: \[ 6.022 \, \times \, 10^{23} \, \text{molecules} / \, \mathrm{mol} \times 1.23 \, \times \, 10^{2} \, \mathrm{g} = 7.40826 \, \times \, 10^{25} \, \text{molecules} \cdot \mathrm{g} / \, \mathrm{mol} \] Now, divide by the denominator: \[ \frac{7.40826 \, \times \, 10^{25} \, \text{molecules} \cdot \mathrm{g} / \, \mathrm{mol}}{46.07 \mathrm{g} / \, \mathrm{mol}} = 1.61 \, \times \, 10^{24} \, \text{molecules} \] The result is \(1.61 \, \times \, 10^{24} \, \text{molecules}\).

- Calculate Part (c)

Identify the given values and plug them into the equation: \[ (6.022 \, \times \, 10^{23} \, \text{atoms} \, / \, \mathrm{mol})(1.28 \, \times \, 10^{-18} \, \mathrm{J} / \, \text{atom}) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] First, calculate \( \frac{1}{2^2} - \frac{1}{3^2} \): \[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \] Then multiply the values: \[ 6.022 \, \times \, 10^{23} \, \text{atoms} / \, \mathrm{mol} \times 1.28 \, \times \, 10^{-18} \, \mathrm{J} / \, \text{atom} \times \frac{5}{36} = 1.07 \, \times \, 10^{5} \, \mathrm{J} / \, \mathrm{mol} \] The result is \(1.07 \, \times \, 10^{5} \, \mathrm{J} / \, \mathrm{mol}\).

Key concepts

Significant Figures

Significant figures are crucial in chemistry calculations because they indicate the precision of measured values. Significant figures include all known digits plus one estimated digit. For example, in the number 6.022, there are four significant figures. When performing calculations, the number of significant figures in the result should be determined by the number with the least significant figures used in the calculation.
Pay attention to the following rules:
• All nonzero digits are significant.
• Any zeros between significant digits are also significant.
• Leading zeros are not significant.
• Trailing zeros in a decimal number are significant.
Always round off your final answer to match the number of significant figures of the least precise number in your calculation.

Unit Conversions

Unit conversions involve changing a measurement from one unit to another. This is essential in chemistry as different units are often used for measuring substances.
Always ensure that your final answer is in the correct unit by using conversion factors. For example, converting grams to kilograms involves dividing by 1,000 because there are 1,000 grams in a kilogram.
Use units in your calculations to help keep track of what you are measuring and to ensure that they cancel out properly, leaving you with the correct unit in the final result.
The key is to multiply the measurement by a conversion factor that relates the two units, and ensures the unwanted units cancel out, leaving the desired units.

Scientific Notation

Scientific notation is a way to express very large or very small numbers more conveniently. It uses powers of ten to simplify these values.
For example, instead of writing 0.00000000006626, you would write 6.626 x 10^-34. This notation makes calculations easier, as you can simply add or subtract exponents when multiplying or dividing numbers in scientific notation.
To convert a number into scientific notation, move the decimal point until you have a number between 1 and 10, then count how many places you moved the decimal point. This count becomes the exponent of 10.
For example, for the number 6.626 x 10^-34:
• The coefficient (6.626) must be a number between 1 and 10.
• The exponent (-34) tells you how many places to move the decimal point to revert to the original number.

Mole Calculations

The mole is a fundamental unit in chemistry that measures the amount of substance. One mole equals 6.022 x 10^23 particles, which can be atoms, molecules, or ions depending on the context.
Mole calculations often involve converting between grams, moles, and number of particles.
Use Avogadro's number (6.022 x 10^23) to convert between moles and number of particles and the molar mass of a substance to convert between grams and moles.
For example:
• To find the number of molecules in a 2-mole sample: 2 moles x 6.022 x 10^23 molecules/mole = 1.2044 x 10^24 molecules.
• To convert grams into moles, use the molar mass of the substance (found on the periodic table): mass (g) / molar mass (g/mol) = moles.
Practicing mole calculations will improve your understanding and efficiency in chemistry problems.