When determining the vapor pressure of water, the bubble volume is 1.2mL at 5 degrees celsius.

To determine the moles of dry air in the bubble at 5 degrees Celsius, we can use the ideal gas law. The ideal gas law equation is given as:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

Given that the bubble volume is 1.2 mL at 5 degrees Celsius, we need to convert the volume to liters and the temperature to Kelvin.

1 mL = 0.001 L
5 degrees Celsius = 5 + 273.15 Kelvin

Substituting the values into the ideal gas law equation:

(1 atm) * (0.001 L) = n * (0.0821 L·atm/(K·mol)) * (278.15 K)

Simplifying the equation:

0.001 atm·L = 0.0821 * n * 278.15 K

To solve for "n" (number of moles), divide both sides of the equation by (0.0821 × 278.15):

n = (0.001 atm·L) / (0.0821 L·atm/(K·mol) × 278.15 K)

n ≈ 0.00004799 moles

Therefore, there are approximately 0.00004799 moles of dry air in the bubble at 5 degrees Celsius.

Now, to determine the partial pressure of the dry air at a higher temperature of 75 degrees Celsius and a bubble volume of 8.6 mL. We can use the same ideal gas law equation and procedure.

Convert 8.6 mL to liters and 75 degrees Celsius to Kelvin:

8.6 mL = 0.0086 L
75 degrees Celsius = 75 + 273.15 Kelvin

Substituting the values into the ideal gas law equation:

P * (0.0086 L) = n * (0.0821 L·atm/(K·mol)) * (348.15 K)

Simplifying the equation:

0.0086 atm·L = 0.0821 * n * 348.15 K

To solve for "P" (partial pressure), divide both sides of the equation by (0.0086 × 0.0821 × 348.15):

P = (0.0086 atm·L) / (0.0821 L·atm/(K·mol) × 348.15 K)

P ≈ 0.0292 atm

Therefore, the partial pressure of the dry air at 75 degrees Celsius and a bubble volume of 8.6 mL is approximately 0.0292 atm.