To convert the enthalpy of vaporization of water from kJ/mol to J/g, you need to divide the given value in kJ/mol by the molar mass of water (18 g/mol). This will give you the enthalpy of vaporization in J/g.
It takes 333.51 or 334 joules to evaporate 1 gram of H2O.
To react with 0.373 mol of ethylene (C2H4), you need an equal number of moles of water (H2O) based on the balanced chemical equation: C2H4 + H2O โ C2H5OH This means you need 0.373 mol of water to react with 0.373 mol of ethylene. To convert moles to grams, you would multiply 0.373 mol by the molar mass of water (18.015 g/mol) to get the grams needed.
To convert 2 lb mol of C3H8 to grams, you need to first determine the molar mass of C3H8 (propane) which is approximately 44.1 g/mol. Then, you can multiply the number of moles (2 lb mol) by the molar mass to get the mass in grams. So, 2 lb mol of C3H8 is equal to 2 lb mol * 44.1 g/mol = 88.2 g.
To convert 1.5 mol of water into grams, you can use the molar mass of water, which is approximately 18.015 g/mol. Multiply 1.5 mol by the molar mass to get the grams of water. Therefore, 1.5 mol of water is equal to 27.02 grams.
There is approximately 0.055 moles of water (H2O) in 1 gram. This can be calculated by dividing the given mass (1 gram) by the molar mass of water (18 grams/mol).
Adding 4 mol sugar to 1 g (gram) water is impossible !
To convert the enthalpy of vaporization of water from kJ/mol to J/g, you need to divide the given value in kJ/mol by the molar mass of water (18 g/mol). This will give you the enthalpy of vaporization in J/g.
It takes 333.51 or 334 joules to evaporate 1 gram of H2O.
There are 1.5 moles of water molecules in a 27 gram sample of water. This is calculated by dividing the mass of the sample (27 grams) by the molar mass of water (18 grams/mol).
To react with 0.373 mol of ethylene (C2H4), you need an equal number of moles of water (H2O) based on the balanced chemical equation: C2H4 + H2O โ C2H5OH This means you need 0.373 mol of water to react with 0.373 mol of ethylene. To convert moles to grams, you would multiply 0.373 mol by the molar mass of water (18.015 g/mol) to get the grams needed.
The molar mass of Li2O is 29.88 g/mol and the molar mass of H2O is 18.02 g/mol. Using stoichiometry, we find that 2.72 grams of Li2O requires 2.72 grams of water to react in a 1:1 ratio based on the balanced chemical equation.
The gram formula mass for H2O is 18.015 g/mol.
3,5810e23 is equal to 0,59 mol.
To convert 2 lb mol of C3H8 to grams, you need to first determine the molar mass of C3H8 (propane) which is approximately 44.1 g/mol. Then, you can multiply the number of moles (2 lb mol) by the molar mass to get the mass in grams. So, 2 lb mol of C3H8 is equal to 2 lb mol * 44.1 g/mol = 88.2 g.
To convert 1.5 mol of water into grams, you can use the molar mass of water, which is approximately 18.015 g/mol. Multiply 1.5 mol by the molar mass to get the grams of water. Therefore, 1.5 mol of water is equal to 27.02 grams.
Exactly the same number of mols. 1 mol of oxigen atoms produces 1 mol of water molecules.