Question

suppose that the rms speed of carbon dioxide molecules, with molar mass of 44.0 g/mol, in a flame is found to be 1.3 × 105 m/s.

Answer 1

The temperature represented by the average velocity of carbon dioxide molecules in a flame can be calculated using the formula T = (m * Urms^2) / (3k), where Urms is the average velocity, k is Boltzmann's constant, and m is the molar mass of the molecule.

Step-by-step explanation:

The question asks for the temperature represented by the average velocity of carbon dioxide molecules in a flame. To find the temperature, we can use the equation: Urms = √(3kT/m). where Urms is the average velocity, k is Boltzmann's constant (1.38 × 10^-23 J/K), T is the temperature in Kelvin, and m is the molar mass of the molecule.

By rearranging the equation, we can solve for T: T = (m * Urms^2) / (3k). Plugging in the values, we have:T = (44.0 g/mol * (1.3 × 10^5 m/s)^2) / (3 * 1.38 × 10^-23 J/K) .Calculating this expression gives us the temperature represented by the average velocity of carbon dioxide molecules in a flame.