Question

1.) A 71.1 mL sample of oxygen was collected over water. The atmospheric pressure is 0.853 atm and the temperature of the water is 19.6 °C. What are the moles of oxygen?

2. A sample of hydrogen peroxide solution has a mass of 4.027 g. After reacting with the enzyme solution, the moles of oxygen are 0.00264 moles. What is the percent hydrogen peroxide in the solution?

Answer 1

(1) We know, PV=nRT Here, P = Pressure = 0.85 3. afm v= volume = 71.1 mL -0.0711 Litre R = 0.082 atmol/imolox T= 19.6°C (19-6+273) k 292.6k ? i n n = mole a PV RT - (0.853 afin)x0.07111) (0.0 0.082 afmet ***/wof. )*(292.6) 2.5 x 10 - 3 mole 0.0025 mole i Moles of oxygen = 0.0025 mole

H₂O2 (2) Enzyme 02 + H₂O Initial): 4.0249 (After (0.00264% 32 Reaction): = 0.089489 /mol of oxygen Remaining H2O2 = (4.027 g 0.08448 3.942 g/mol 17/mol produced . 3. percent of H₂O2 in solution - 3.942 g 4.027 g * 100 % 97.9% 0 Answer