To find the amount of oxygen required for the combustion of liquid benzene, you would need to balance the combustion reaction equation for benzene. Calculate the moles of benzene from the given mass, then use the stoichiometry of the balanced equation to determine the moles of oxygen required. Finally, convert the moles of oxygen to liters using the ideal gas law.
To find the volume of a liquid in liters, you can use a measuring container marked in liters. Simply pour the liquid into the container and read the volume indicated on the container in liters.
Liters liquid 1000ml/1L g/ml mol/g Hfusion
The balanced chemical equation for the combustion of butane is: 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O. This means it requires 13 moles of O₂ for every 2 moles of butane. The volume of O₂ needed would be (14.9 L * 13 mol) / 2 = 96.85 L of O₂.
half a liter
Approximately 860 liters of gaseous oxygen can be obtained from 1 cubic foot of liquid oxygen when it vaporizes. This expansion occurs due to the change in state from liquid to gas.
Molarity is a concentration unit of solutions, pure benzene is not a solution so this term is meaningless for a pure liquid. In other words: Molarity of a solution is a numerical way of saying exactly how much solute is dissolved in a solvent .Molarity is equal to the moles of solute divided by the liters of solution. Molarity of a solution can be calculated by using the following formula: M= moles of solute/liters of solution
Liters liquid 1000ml/1L g/ml mol/g Hfusion
To determine the volume of oxygen required for the combustion of 11 liters of ethylene (C₂H₄) at standard temperature and pressure (STP), we first note that the balanced combustion reaction is: C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O. This indicates that 1 mole of ethylene requires 3 moles of oxygen. Since 11 liters of ethylene corresponds to approximately 11 moles at STP, the oxygen needed would be 3 times that volume, resulting in 33 liters of oxygen at STP.
To determine the volume of oxygen gas required to react with 1.18 mol of benzene (C6H6), we first need the balanced chemical equation for the complete combustion of benzene, which is: [ \text{C}_6\text{H}_6 + 7 \text{O}_2 \rightarrow 6 \text{CO}_2 + 3 \text{H}_2\text{O} ] From the equation, 1 mole of benzene reacts with 7 moles of oxygen. Therefore, for 1.18 moles of benzene, the required moles of oxygen are: [ 1.18 , \text{mol C}_6\text{H}_6 \times 7 , \text{mol O}_2/\text{mol C}_6\text{H}_6 = 8.26 , \text{mol O}_2 ] At 0 °C and 1 ATM, 1 mole of gas occupies 22.4 liters, so the volume of oxygen needed is: [ 8.26 , \text{mol O}_2 \times 22.4 , \text{L/mol} \approx 185.86 , \text{L} ] Thus, approximately 185.86 liters of oxygen gas is required.
To find the volume of a liquid in liters, you can use a measuring container marked in liters. Simply pour the liquid into the container and read the volume indicated on the container in liters.
1 liquid oz - 0.0295735296 liters.
579 mL = 0.579 liters.
no in pounds or kilos liters are for liquid
Liters liquid 1000ml/1L g/ml mol/g Hfusion
On average, a gallon of gasoline contains around 0.6-2% benzene by volume, which is roughly 0.05-0.15 gallons or 0.2-0.6 liters of benzene per gallon of gasoline. Benzene is a naturally occurring component of crude oil and is present in small amounts in gasoline.
No. Five liters is equal to about 1.32 US gallons of liquid.
The balanced chemical equation for the combustion of butane is: 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O. This means it requires 13 moles of O₂ for every 2 moles of butane. The volume of O₂ needed would be (14.9 L * 13 mol) / 2 = 96.85 L of O₂.