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What 6 moles of a compound produce 84 J of energy what is the Hreaction in Jmol?
-14 J/mol
If 3 moles of a compound use 24 J of energy ina reaction what is the Hreaction in Jmol?
To find the enthalpy change (ΔH) for the reaction in J/mol, you divide the total energy used by the number of moles. Given that 3 moles of the compound use 24 J of energy, you calculate ΔH as follows: ΔH = 24 J / 3 moles = 8 J/mol. Thus, the enthalpy change for the reaction is 8 J/mol.
If 6 moles of a compound produce 84 J of energy what is the change of H reaction in Jmol?
To find the change in enthalpy (ΔH) for the reaction in J/mol, divide the total energy produced by the number of moles. For 6 moles producing 84 J, ΔH = 84 J / 6 moles = 14 J/mol. Thus, the change in enthalpy for the reaction is 14 J/mol.
If 3 moles of a compound use 24 J of energy in a reaction what is the triangle H reaction in Jmol?
To find the enthalpy change (ΔH) per mole, divide the total energy used by the number of moles. In this case, 24 J divided by 3 moles equals 8 J/mol. Therefore, the ΔH for the reaction is 8 J/mol.
If 6 moles of a compound produce 84 j of energy what is the deltaHreaction jmol?
To find the enthalpy change (( \Delta H )) per mole of the compound, divide the total energy produced by the number of moles. In this case, ( \Delta H = \frac{84 , \text{J}}{6 , \text{moles}} = 14 , \text{J/mol} ). Therefore, the ( \Delta H ) for the reaction is 14 J/mol.
If 3 moles of a compound use 24 J of energy in a reaction what is the Hreaction in Jmol A. -8 Jmol?
The enthalpy change for the reaction would be -8 J/mol, as it is the energy change per mole of the compound reacted.
If 6 moles of a compound produce 84 J of energy what is the Hreaction in Jmol?
The heat of reaction per mole can be calculated by dividing the energy produced by the number of moles. In this case, 84 J of energy produced by 6 moles of the compound gives a heat of reaction of 14 J/mol.
What 6 moles of a compound produce 84 J of energy what is the Hreaction in Jmol?
-14 J/mol
If 3 moles of a compound use 24 J of energy ina reaction what is the Hreaction in Jmol?
To find the enthalpy change (ΔH) for the reaction in J/mol, you divide the total energy used by the number of moles. Given that 3 moles of the compound use 24 J of energy, you calculate ΔH as follows: ΔH = 24 J / 3 moles = 8 J/mol. Thus, the enthalpy change for the reaction is 8 J/mol.
If 6 moles of a compound produce 84 J of energy what is the H reaction in J mol?
-14 J/Mol
If 3 moles of a compound use 12 J of energy in a reaction what is the H reaction in kJmol?
12 J/3 moles = 4 J/mole. Thus, H of reaction in kJ/mole = 0.004 kJ/mole
If 6 moles of a compound produce 84 j of energy what is the H reaction in jmol?
84 J/6 moles = 14 J/mole = ∆H
If 6 moles of a compound produce 84 J of energy what is the change of H reaction in Jmol?
To find the change in enthalpy (ΔH) for the reaction in J/mol, divide the total energy produced by the number of moles. For 6 moles producing 84 J, ΔH = 84 J / 6 moles = 14 J/mol. Thus, the change in enthalpy for the reaction is 14 J/mol.
If 3 moles of a compound use 24 J of energy in a reaction what is the triangle H reaction in Jmol?
To find the enthalpy change (ΔH) per mole, divide the total energy used by the number of moles. In this case, 24 J divided by 3 moles equals 8 J/mol. Therefore, the ΔH for the reaction is 8 J/mol.
If 6 moles of a compound produce 84 J of energy what is the delta H reaction in jmol?
The enthalpy change (ΔH) per mole can be found by dividing the energy produced by the moles of the compound. In this case, ΔH = 84 J / 6 mol = 14 J/mol. Therefore, the enthalpy change per mole of the compound is 14 J/mol.
If 6 moles of a compound produce 84 j of energy what is the deltaHreaction jmol?
To find the enthalpy change (( \Delta H )) per mole of the compound, divide the total energy produced by the number of moles. In this case, ( \Delta H = \frac{84 , \text{J}}{6 , \text{moles}} = 14 , \text{J/mol} ). Therefore, the ( \Delta H ) for the reaction is 14 J/mol.
IF a reaction vessel contains 0.15 moles of LiOH and 0.08 moles of CO2 which compound is the limiting reagent?
Carbon dioxide is the limiting reagent.
