16.05g/mol
16.0425g/mole
The standard enthalpy of combustion for methane is -890 kJ/mol.
Taking carbon @ 12.01 g/mol & hydrogen @ 1.01 g/ mol methane has 1 carbon & 4 hydrogen, so it is:12.01g + 4(1.01g) = 16.05g per mole.For 4 moles: 4 x 16.05 g = 64.20 g
Find out the percentage of hydrogen in the molar mass of methane. Molar mass of CH4: C = 1 * 12.01 g = 12.01 g H = 4 * 1.01 g = 4.04 g Total = 16.05 g 4.04 g/16.05 g * 100% = 25.171% 0.25171 * 20 g = 5.0342 g There are about 5.03 grams of hydrogen in 20 grams of methane gas.
1. Find the molar mass of the hydrate (Calcium Chloride Dihydrate).Find the molar mass of water and the anhydrate (anhydrate + water = hydrate); add the molar mass values of each to find the molar mass of the hydrate.Molar Mass CaCl2: 110.98g+ Molar Mass H2O: 36.04g*Molar Mass CaCl2 * 2H2O: 147.01gFinding Molar Mass# atoms element A * atomic mass element A = Mass A# atoms element B * atomic mass element B = Mass B... etc.Add up all the mass values and you have the value for molar mass. Do this for both the anhydrate and the water molecules. Add these values together to find the molar mass of the hydrate.Molar Mass Anhydrate + Molar Mass Water Molecules* = Molar Mass Hydrate* Tip: the molar mass of water for all hydrate calculations is 18.02g x number of water molecules. This number may be useful to remember on the day of the test or while doing practice problems.*2. Calculate the percentage of water in hydrate.Divide the molar mass of water by the molar mass of the hydrate, and multiply result by 100%.36.04g147.01g x 100%Percent water in hydrate is 24.52%.
16.05 g/mol
16.0425g/mole
The standard enthalpy of combustion for methane is -890 kJ/mol.
The molar heat of combustion of methane (890 kJ/mol) is higher than that of water gas (525 kJ/mol)
Taking carbon @ 12.01 g/mol & hydrogen @ 1.01 g/ mol methane has 1 carbon & 4 hydrogen, so it is:12.01g + 4(1.01g) = 16.05g per mole.For 4 moles: 4 x 16.05 g = 64.20 g
Find out the percentage of hydrogen in the molar mass of methane. Molar mass of CH4: C = 1 * 12.01 g = 12.01 g H = 4 * 1.01 g = 4.04 g Total = 16.05 g 4.04 g/16.05 g * 100% = 25.171% 0.25171 * 20 g = 5.0342 g There are about 5.03 grams of hydrogen in 20 grams of methane gas.
Methane reacts with oxygen in the following way. CH4 + 3 O2 --> CO2 + 4 H2O. If 5 moles of oxygen react with 2.8 moles of methane, only 1.67 moles of methane would be consumed because of the molar ratio 1:3. This would produce 1.67 moles of carbon dioxide and 6.67 moles of water.
1. Find the molar mass of the hydrate (Calcium Chloride Dihydrate).Find the molar mass of water and the anhydrate (anhydrate + water = hydrate); add the molar mass values of each to find the molar mass of the hydrate.Molar Mass CaCl2: 110.98g+ Molar Mass H2O: 36.04g*Molar Mass CaCl2 * 2H2O: 147.01gFinding Molar Mass# atoms element A * atomic mass element A = Mass A# atoms element B * atomic mass element B = Mass B... etc.Add up all the mass values and you have the value for molar mass. Do this for both the anhydrate and the water molecules. Add these values together to find the molar mass of the hydrate.Molar Mass Anhydrate + Molar Mass Water Molecules* = Molar Mass Hydrate* Tip: the molar mass of water for all hydrate calculations is 18.02g x number of water molecules. This number may be useful to remember on the day of the test or while doing practice problems.*2. Calculate the percentage of water in hydrate.Divide the molar mass of water by the molar mass of the hydrate, and multiply result by 100%.36.04g147.01g x 100%Percent water in hydrate is 24.52%.
you would have to know the following values, Absorbance (A), Concentration (C), and cell length (l) and plug it into the formula A=eCl or C = A/Cl
False, I'd think.
C2H6 is a gas (in chemistry), similar to methane, propane, and butane. I'm not sure what the name of this gas is, but it ends in -ane. Sure of that.
Water is polar and has a bent shape. Methane is non-polar and is tetrahedral. These molecular-level differences help explain why, despite their similar molar masses, they have vastly different physical properties.