ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction? First you need to write a balanced chemical equation.
Weight of the balloon = 2.0 g
Six weights each of mass 30.0 g is added to the balloon.
Total mass of the balloon = 2.0 g + 6*30.0 g = 182 g
Density of salt water = 1.02 g/mL
Calculating the volume from mass and density:
Converting the volume from mL to cubic cm:
Assuming the balloon to be a sphere,
Volume of the sphere = 
π
r = 3.49 cm
Radius of the balloon = 3.49 cm
Diameter of the balloon = 2 r = 2*3.49 cm = 6.98cm
To start, 1 cubic centimeter = 1 milliliter, so now you have 1.11g/mL.
Now multiply 1.11 by 387 to get the mass of antifreeze in grams, since the mL is canceled out.
387 mL x 1.11g/mL = 429.57 g
