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.
Answer:
713.51 N/m
Explanation:
Hook's Law: This law states that provided the elastic limit is not exceeded, the extension in an elastic material is directly proportional to the applied force.
From hook's law,
F = ke ...........................Equation 1
Where F = Force exerted on the bowstring, e = Extension/compression of the bowstring, k = Spring constant of the bow.
Make k the subject of the equation,
k = F/e ............................ Equation 2
Given: F = 264 N, e = 0.37 m.
Substitute into equation 2
k = 264/0.37
k = 713.51 N/m
Hence the spring constant of the bow = 713.51 N/m
Use the density to convert volume into mass.
since the density is in g/ml and the volume was given in Liters, we need to first convert the Liters into mililiters. just multiply by 1000 or move the decimal three times.
0.1200 Liters= 120.0 mL
120.0 mL (0.8787 grams/ 1 mL)= 105 grams
1) To find the change in enthalpy, determine the difference between the potential energy of the products and the potential energy of the reactants. (on this diagram, C-A) To find the activation energy, find the difference between the potential energy of the reactants and the "peak" of the curve (on this diagram, B-A). For this diagram, both the enthalpy and activation energy are positive.
2) If the reaction was exothermic, enthalpy would be negative, and the potential energy of the reactants would be greater than the potential energy of the products.
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