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117(g K) * [1.008(g/mol H) / 39.098(g/mol K)] = 3.02 gram hydrogen
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
4.50gramH20
Balanced equation first. 2H2 + O2 -> 2H2O 250 moles O2 (2 mole H2O/1 mole O2) = 500 mole H2O produced now, since I am forgetful, I will use density formula at 25 C Density = grams/milliliters 500 moles H2O (18.016 grams/1 mole H2O) = 9008 grams H2O 0.9982 g/ml = 9008 grams/milliliters 9024.24 milliliters H2O this is...... 9.02 liters of water produced in this reaction.
This oxide may be CdO.
60 grams.
117(g K) * [1.008(g/mol H) / 39.098(g/mol K)] = 3.02 gram hydrogen
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
4.50gramH20
Balanced equation first. 2H2 + O2 -> 2H2O 250 moles O2 (2 mole H2O/1 mole O2) = 500 mole H2O produced now, since I am forgetful, I will use density formula at 25 C Density = grams/milliliters 500 moles H2O (18.016 grams/1 mole H2O) = 9008 grams H2O 0.9982 g/ml = 9008 grams/milliliters 9024.24 milliliters H2O this is...... 9.02 liters of water produced in this reaction.
This oxide may be CdO.
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
88
Liters can't be converted to grams. Liters measure volume, while grams measure mass.
The answer is 900 kg.
Liters can't be converted to grams. Liters measure volume, while grams measure mass.
3.61g/L D=m/v