Question

A sample of carbon monoxide initially at C was heated to 1.90 times 10²ᶜircC. If the volume of the carbon monoxide sample is 908.6mL at 1.90 times 10²ᶜircC, what was its volume at 95.0 C?

a. 750.2mL
b. 782.4mL
c. 815.7mL
d. 849.1mL

Answer 1

The volume of the carbon monoxide sample at 95.0°C is approximately 4569.8 mL.

Step-by-step explanation:

To find the volume of the carbon monoxide sample at 95.0°C, we can use the formula:

V₁ / T₁ = V₂ / T₂

Where V₁ is the initial volume, T₁ is the initial temperature, V₂ is the final volume, and T₂ is the final temperature.

Plugging in the given values:

908.6 mL / (1.90 × 10²°C) = V₂ / (95.0°C)

Solving for V₂:

V₂ = 908.6 mL × (95.0°C) / (1.90 × 10²°C) = 4569.8 mL

Therefore, the volume of the carbon monoxide sample at 95.0°C is approximately 4569.8 mL.