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How many moles are in 0.289 g of methane?
To find the number of moles in 0.289 g of methane, first calculate the molar mass of methane (CH4), which is approximately 16 g/mol. Next, divide the given mass (0.289 g) by the molar mass to obtain the number of moles. Therefore, 0.289 g of methane is equivalent to 0.289 g / 16 g/mol ≈ 0.018 moles of methane.
Which of the following shows that the combustion of methane produces 802 kJ mol of energy?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ/mol
What is an example that shows that the combustion of methane produces 802 kJmol of energy?
The enthalpy of combustion is determined by calorimetry.
How much heat is released when methane and oxygen are burned?
890 kJ of energy are released when I mole of methane (16 g) is completely burned in oxygen.
How many moles of carbon dioxide are produced from the complete combustion of 100.0 grams of methane?
To determine the moles of carbon dioxide produced from the combustion of methane, we first need to balance the chemical equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. From the balanced equation, we see that 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is 16.05 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 100.0 grams of methane is equivalent to 100.0 g / 16.05 g/mol = 6.23 moles of methane, which would produce 6.23 moles of carbon dioxide.
How many moles are in 0.289 g of methane?
To find the number of moles in 0.289 g of methane, first calculate the molar mass of methane (CH4), which is approximately 16 g/mol. Next, divide the given mass (0.289 g) by the molar mass to obtain the number of moles. Therefore, 0.289 g of methane is equivalent to 0.289 g / 16 g/mol ≈ 0.018 moles of methane.
Which of the following shows that the combustion of methane produces 802 kJ mol of energy?
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ/mol
What is an example that shows that the combustion of methane produces 802 kJmol of energy?
The enthalpy of combustion is determined by calorimetry.
How much heat is released when methane and oxygen are burned?
890 kJ of energy are released when I mole of methane (16 g) is completely burned in oxygen.
How many moles of carbon dioxide are produced from the complete combustion of 100.0 grams of methane?
To determine the moles of carbon dioxide produced from the combustion of methane, we first need to balance the chemical equation for the combustion of methane: CH4 + 2O2 → CO2 + 2H2O. From the balanced equation, we see that 1 mole of methane produces 1 mole of carbon dioxide. The molar mass of methane (CH4) is 16.05 g/mol, and the molar mass of carbon dioxide (CO2) is 44.01 g/mol. Therefore, 100.0 grams of methane is equivalent to 100.0 g / 16.05 g/mol = 6.23 moles of methane, which would produce 6.23 moles of carbon dioxide.
If in the previous problem only 25.9 grams of water vapor were formed how much methane actually reacted with oxygen?
The equation for the reaction between methane (CH4) and oxygen is: CH4 + 2O2 → CO2 + 2H2O From the previous information, if 25.9 grams of water vapor were formed, this corresponds to 25.9 grams ÷ 18.0 g/mol = 1.44 moles of water. This means 0.72 moles of methane reacted. If the molar mass of methane is 16.0 g/mol, then 0.72 moles of methane corresponds to 0.72 moles × 16.0 g/mol = 11.52 grams of methane reacted with oxygen.
The formula for methane is CH4. Four moles of methane would have a mass of grams.?
To calculate the mass of four moles of methane (CH4), first determine the molar mass of methane. The molar mass of carbon (C) is approximately 12.01 g/mol, and hydrogen (H) is about 1.01 g/mol. Thus, the molar mass of CH4 is 12.01 + (4 × 1.01) = 16.05 g/mol. Therefore, the mass of four moles of methane is 4 moles × 16.05 g/mol = 64.2 grams.
The formula for methane is CH4. Four moles of methane would have a mass of how many grams?
To calculate the mass of four moles of methane (CH4), we first need to determine the molar mass of methane. The molar mass of CH4 is the sum of the atomic masses of carbon (12.01 g/mol) and hydrogen (1.008 g/mol) multiplied by the number of atoms in the compound. Therefore, the molar mass of CH4 is 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol. To find the mass of four moles of methane, we multiply the molar mass by the number of moles: 16.04 g/mol x 4 mol = 64.16 grams. Therefore, four moles of methane would have a mass of 64.16 grams.
How many liters of water vapor at STP are produced by the combustion of 12.0g of methane?
To determine the liters of water vapor produced by the combustion of 12.0 g of methane (CH₄), we first calculate the moles of methane using its molar mass (approximately 16.04 g/mol). The balanced equation for the combustion of methane is: CH₄ + 2 O₂ → CO₂ + 2 H₂O. Thus, 1 mole of methane produces 2 moles of water vapor. Therefore, 12.0 g of methane corresponds to about 0.748 moles, resulting in approximately 1.496 moles of water vapor, which at STP (22.4 L/mol) corresponds to roughly 33.5 liters of water vapor.
How many liters of oxygen are needed to exactly react with 27.8 g of methane at STP?
1) First find the number of moles of methane in 27.8 g using the molar mass.See the Related Question to the left of this answer "How do you convert from grams to moles and also from moles to grams?" to do that.2) Then write the balanced reaction. Methane (CH4) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). See the related question "How do you balance a chemical reaction?" to do that.3) That will tell you the ratio of moles of methane to moles of oxygen (it will be 2 to 1). So from Part 1, multiply the number of moles of methane by 2 to get moles of oxygen. Then, use the Ideal Gas Law to find out how many liters that will take up at STP. Use the Related Question link "How do you solve Ideal Gas Law problems?" to do that.
How many moleclues of water will be produced if 13.15g of methane is burned?
To determine the number of water molecules produced from the combustion of 13.15 g of methane (CH₄), we first write the balanced equation for the reaction: CH₄ + 2 O₂ → CO₂ + 2 H₂O. The molar mass of methane is approximately 16.04 g/mol. Therefore, 13.15 g of methane corresponds to about 0.819 moles of CH₄, which will produce 1.638 moles of water (H₂O) since each mole of methane produces 2 moles of water. Multiplying 1.638 moles by Avogadro's number (6.022 x 10²³) yields approximately 9.85 x 10²³ molecules of water.
What are the steps needed to convert 77.2 grams of methane to molecules of methane. Please wrote them out ( indluding tchart)?
Step1: Find molar mass of methane, CH4. C=12; H=1; 12+1+1+1+1 = 16 g/moleStep2: Convert grams to moles. 77.2 g x 1 mole/16 g = 4.825 moles methaneStep3: Convert moles to molecules using Avogadro's number.4.825 moles x 6.02x10^23 molecules/mole = 2.90x10^24 moleculesUnfortunately, I don't know what a tchart is, so I didn't include it.
