Question

Carry out each calculation, paying special attention to significant figures, rounding, and units: (a) \(\frac{4.32 \times 10^{7} \mathrm{~g}}{\frac{4}{3}(3.1416)\left(1.95 \times 10^{2} \mathrm{~cm}\right)^{3}}\) (The term \(\frac{4}{3}\) is exact.) (b) \(\frac{\left(1.84 \times 10^{2} \mathrm{~g}\right)(44.7 \mathrm{~m} / \mathrm{s})^{2}}{2}\) (The term 2 is exact.) (c) \(\frac{\left(1.07 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\right)^{2}\left(3.8 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\right)}{\left(8.35 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\right)\left(1.48 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\right)^{3}}\)

Short answer

(a) 1.40 g/cm³(b) 1.84 x 10⁵ g m²/s²(c) 13.5

Step-by-step solution

Step 1a: Write Down the Given Values for Part (a)

Identify and write down the given values: Mass, \(m = 4.32 \times 10^{7} \text{ g}\)Radius, \(r = 1.95 \times 10^{2} \text{ cm}\)Constant, \( \frac{4}{3} \times \text{Pi} \times r^{3}\)

Step 2a: Convert Units if Necessary

Since all units are consistent (grams and centimeters), there's no need for conversion in this problem.

Step 3a: Calculate the Volume

Using \(V = \frac{4}{3} \times \text{Pi} \times r^{3}\), substitute \( \text{Pi} = 3.1416\) and \(r = 1.95 \times 10^{2} \text{ cm}\): \(V = \frac{4}{3} \times 3.1416 \times (1.95 \times 10^{2})^{3}\)\(V = \frac{4}{3} \times 3.1416 \times (7.41 \times 10^{6})\) \(V = 3.09 \times 10^{7} \text{ cm}^3\)

Step 4a: Calculate the Density

Density \( \rho = \frac{m}{V} \)\( \rho = \frac{4.32 \times 10^{7} \text{ g}}{3.09 \times 10^{7} \text{ cm}^{3}} \)\( \rho = 1.40 \text{ g/cm}^{3}\)

Step 1b: Write Down the Given Values for Part (b)

Identify and write down the given values: Mass, \(m = 1.84 \times 10^{2} \text{ g}\)Velocity, \(v = 44.7 \text{ m/s}\)Constant, \(2\)

Step 2b: Calculate the Kinetic Energy

Using \( KE = \frac{1}{2} \times m \times v^{2} \), substitute \(m = 1.84 \times 10^{2} \text{ g}\) and \(v = 44.7 \text{ m/s}\):\( KE = \frac{1.84 \times 10^{2} \text{ g}}{2} \times (44.7 \text{ m/s})^{2}\)\( KE = 0.92 \times 10^{2} \times 1.998 \times 10^{3} \text{ g m}^{2}/s^{2}\)\( KE = 1.84 \times 10^{5} \text{ g m}^{2}/s^{2}\)

Step 1c: Write Down the Given Values for Part (c)

Identify and write down the given values: \([A] = 1.07 \times 10^{-4} \text{ mol/L}\)\([B] = 3.8 \times 10^{-3} \text{ mol/L}\)\([C] = 8.35 \times 10^{-5} \text{ mol/L}\)\([D] = 1.48 \times 10^{-2} \text{ mol/L}\)

Step 2c: Apply the Reaction Quotient Formula

For the reaction quotient \( Q\), apply the formula:\( Q = \frac{[A]^2 [B]}{[C][D]^3}\)Substitute the values:\( Q = \frac{(1.07 \times 10^{-4})^2 (3.8 \times 10^{-3})}{(8.35 \times 10^{-5}) (1.48 \times 10^{-2})^3}\)

Step 3c: Simplify the Expression

Calculating each part: \((1.07 \times 10^{-4})^2 = 1.1449 \times 10^{-8}\)\((1.48 \times 10^{-2})^3 = 3.24 \times 10^{-6}\)Then substitute back\( Q = \frac{1.1449 \times 10^{-8} \times 3.8 \times 10^{-3}}{8.35 \times 10^{-5} \times 3.24 \times 10^{-6}}\)\( Q = 1.35 \times 10^{1}\)

Key concepts

density calculation

Density is an essential concept in chemistry that describes how much mass is contained in a given volume. The formula to calculate density is:
\( \rho = \frac{m}{V} \)
where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.

To understand how to use this formula, let's go through the given example from the exercise:

• Identify and write down the given values: Mass \( m = 4.32 \times 10^{7} \text{ g} \), Radius \( r = 1.95 \times 10^{2} \text{ cm} \)
• Use the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \)
• Substitute \( \pi = 3.1416 \) and \( r = 1.95 \times 10^{2} \text{ cm} \)
• Calculate the volume: \( V = \frac{4}{3} \times 3.1416 \times (1.95 \times 10^{2})^3 = 3.09 \times 10^{7} \text{ cm}^3 \)
• Finally, solve for density: \( \rho = \frac{m}{V} = \frac{4.32 \times 10^{7} \text{ g}}{3.09 \times 10^{7} \text{ cm}^3} = 1.40 \text{ g/cm}^3 \)

Remember to always pay attention to significant figures in your calculations. In this case, the mass and volume were provided to three significant figures.

kinetic energy formula

Kinetic energy is the energy an object possesses due to its motion, and it's calculated using the formula:
\( KE = \frac{1}{2}mv^2 \)
where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity.

Let's go through the given example:

• Identify the provided values: Mass \( m = 1.84 \times 10^{2} \text{ g} \) and Velocity \( v = 44.7 \text{ m/s} \)
• Use the kinetic energy formula: \( KE = \frac{1}{2}mv^2 \)
• Substitute the values: \( KE = \frac{1.84 \times 10^{2} \text{ g}}{2} \times (44.7 \text{ m/s})^2 \)
• Calculate the kinetic energy: \( KE = 0.92 \times 10^{2} \times 1.998 \times 10^{3} \text{ g m}^2/\text{s}^2 = 1.84 \times 10^{5} \text{ g m}^2/\text{s}^2 \)

It's important to ensure all units are correctly handled and matched. Here, we should ideally convert grams into kilograms to match the standard unit of kinetic energy \( \text{J} \). Always check the units and significant figures in your results.

reaction quotient

The reaction quotient (\( Q \)) is a measure used to determine the direction a reaction will proceed to reach equilibrium. It is calculated using the concentrations of the reactants and products. The formula is:
\( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

Let's go through the provided example:

• Identify and write down the values: \( [A] = 1.07 \times 10^{-4} \text{ mol/L} \), \( [B] = 3.8 \times 10^{-3} \text{ mol/L} \), \( [C] = 8.35 \times 10^{-5} \text{ mol/L} \), and \( [D] = 1.48 \times 10^{-2} \text{ mol/L} \)
• Apply the reaction quotient formula:
  \( Q = \frac{[A]^2 [B]}{[C][D]^3} \)
• Substitute the values: \( Q = \frac{(1.07 \times 10^{-4})^2 (3.8 \times 10^{-3})}{(8.35 \times 10^{-5}) (1.48 \times 10^{-2})^3} \)
• Simplify the exponents and terms: \( (1.07 \times 10^{-4})^2 = 1.1449 \times 10^{-8} \) and \( (1.48 \times 10^{-2})^3 = 3.24 \times 10^{-6} \)
• Calculate the final quotient: \( Q = \frac{1.1449 \times 10^{-8} \times 3.8 \times 10^{-3}}{8.35 \times 10^{-5} \times 3.24 \times 10^{-6}} = 1.35 \times 10^{1} \)

Understanding how to properly use and calculate the reaction quotient will help in predicting whether a reaction will proceed forward or in reverse.