A badly tuned automobile engine can release about 50 moles of carbon monoxide per hour. At 35 ◦C, what volume of carbon monoxide is released in a six-hour period if the atmospheric pressure is 740 Torr?

7.79 × 10³ L

Explanation:

Step 1: Calculate the total number of moles of carbon dioxide released

50 moles of carbon dioxide are released per hour. The moles released in a six-hour period is:

6 h × 50 mol/1 h = 300 mol

Step 2: Convert the temperature to the Kelvin scale

When working with gases, we need to consider the absolute temperature. We can convert from Celsius to Kelvin using the following expression.

K = °C + 273.15

K = 35 °C + 273.15 = 308 K

Step 3: Convert the pressure to atm

We will use the conversion factor 1 atm = 760 Torr.

740 Torr × 1 atm/760 Torr = 0.974 atm

Step 4: Calculate the volume of the gas

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 300 mol × (0.0821 atm.L/mol.K) × 308 K / 0.974 atm

V = 7.79 × 10³ L

a non redox reaction occurs when there is not a transfer of electrons. for redox reactions, which are reduction-oxidation reactions, a change in the oxidation number occurs. elements that are reduced gain electrons, and elements that are oxidized lose electrons.

explanation: