Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.
Regard the principle of utilization of two gas.
Make a consistent control of hardware containing gas.
Make a consistent control of weight diminishing valves giving gas.
No smoking zone.
E = mc^2
E = 8.90 * 10^12 Joules
c = 3 * 10^8 m/s
m = ????
8.90 * 10^12 = m * (3 * 10^8)^2
8.90 * 10^12 = m * 9 * 10^16
9.889 * 10^-4 kg = m <<<<< answer
