If the barometric pressure that day was 75.1 cm Hg, the temperature was 27.0°C and the volume in the pipette decreases from 51.0 mL to 42.0 mL, how many millimoles of gas (total) were removed from the initial air sample? (Enter as a decimal value, not in scientific notation and enter the value only. Note that a millimole is 10-3 mole.)

The total number of mole of the gas that were removed from the initial sample is 0.36 millimole

From the question given above, the following data were obtained:

Pressure (P) = 75.1 cm Hg = 75.1 / 76 = 0.988 atm

Temperature (T) = 27 °C = 27 + 273 = 300 K

Volume (V) = 51 – 42 = 9 mL = 9 / 1000 = 0.009 L

Gas constant (R) = 0.0821 atm.L/Kmol

Using the ideal gas equation (PV = nRT), we shall determine the total number of mole of the gas that were removed from the initial sample. This is illustrated below:

0.988 × 0.009 = n × 0.0821 × 300

0.008892 = n × 24.63

Divide both side by 24.63

n = 0.008892 / 24.63

n = 0.00036 mole

Multiply by 1000 to express in millimole

n = 0.00036 × 1000

Thus, the total number of mole of the gas that were removed from the initial sample is 0.36 millimole

Learn more: link

0.36 millimoles

Explanation:

According to ideal gas equation:

P = pressure of gas = 75.1 cm Hg = 0.98 atm    (1cmHg = 10 mm Hg)

V = Volume of gas = (51.0-42.0) ml = 9.00 ml = 0.009 L

n = number of moles = ?

R = gas constant =

T = Temperature =

Thus 0.36 millimoles were removed from initial air sample

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

i believe the answer is true.