IGNORE THE ANSWERS IN ALREADY PLEASE ANSWER ASAP::: A student performed the expe
ID: 895069 • Letter: I
Question
IGNORE THE ANSWERS IN ALREADY PLEASE ANSWER ASAP:::
A student performed the experiment as described in the lab manual, using 5.00 mL of an aqueous hydrogen peroxide solution, with a density of 1.01 g/mL. The water temperature was 294 K and the barometric pressure was 31.50 in. Hg. After the student immersed the yeast in the peroxide solution, they observed a 43.34 mL change in system volume. The barometric pressure, in torr, is Blank 1 800.0 torr. (1 dec places) The water vapor pressure at the water temperature is Blank 2 18.7 torr. (1 dec place) The pressure exerted by the collected oxygen gas at the water temperature is Blank 3 781.4 torr (1 dec place) or Blank 4 1.03 atm (2 dec places). The volume of collected oxygen gas is Blank 5 .04334 L (4 significant figures). Using R = 0.08206, the number of moles of collected oxygen gas is Blank 6 1.4 (2 dec places) x 10Blank 7 2.8 (integer) mol. So, based upon the number of moles of oxygen gas collected, the number of moles of peroxide in the original 5.00 mL sample is Blank 8 1.4 (2 dec places) x 10Blank 9 2.8 (integer) moles. So, the mass of peroxide (MW = 34.02), is Blank 10 95.256 g (3 dec places)
Explanation / Answer
The water temperature was 294 K and the barometric pressure was 31.50 in. Hg. After the student immersed the yeast in the peroxide solution, they observed a 43.34 mL change in system volume. The barometric pressure, in torr, is 800.0 torr
The water vapor pressure at the water temperature is 18.7 torr
The pressure exerted by the collected oxygen gas at the water temperature is 781.4 torr or 1.03 atm .
The volume of collected oxygen gas is 0.04334 L
the number of moles of collected oxygen gas is 1.40 x 10 = 14 mol.
So, based upon the number of moles of oxygen gas collected, the number of moles of peroxide in the original 5.00 mL sample = 1.4 mol * 10 = 14 moles.
So, the mass of peroxide (MW = 34.02) is 95.256 g