40.79 H2O kj/moles x 1 moles/18 grams= 2.266/1000=0.002266 joules/grams
The heat of solution for barium chloride is -61.7 kJ/mol. To convert this to joules, multiply by 1000 to get -61,700 J/mol.
The first step is to calculate how many moles of Na are present in 1 mg (0.001 g) of Na. Since the molar mass of Na is 22.98977 g/mol, there are 0.000043 moles of Na. Next, convert the ionization energy to joules by multiplying by 1000 (1 kJ = 1000 J) to get 495,800 J/mol. Finally, multiply the number of moles by the energy per mole to find the total energy required: 495,800 J/mol * 0.000043 mol = 21.3 J.
The units for standard Gibbs free energy are joules per mole (J/mol) or kilojoules per mole (kJ/mol).
2.5 g 1 mol/18.02 g (-285.83) kJ/mol
The Gibbs free energy change is calculated from the expression Δ G = Δ H - T(Δ S) For the combustion of ethene (assuming it takes place at 25oC): C2H4 + 3O2 --> 2CO2 + 2H2O you need to find the enthalpy and entropy changes, which are Δ H (combustion) = - 1400 kJ/mol Δ S (combustion) = - 1102 J/mol/K Substituting into the first equation, remembering to divide the entropy value by 1000 because it's in J per mol per kelvin, not kJ, and converting the 25 degrees C to kelvin, we get: Δ G = -1314.35 kJ http://www.docbrown.info/page07/delta3SGc.htm
To convert from kilojoules per mole to joules per gram, you need the molar mass of the substance. Once you have the molar mass, you can convert as follows: 1 kJ/mol = 1000 J/mol 40.79 kJ/mol = 40.79 * 1000 J/mol = 40790 J/mol Then, divide by the molar mass in grams/mol to get joules per gram.
Lets say, for example the enthalpy is equal to 1200 joules/gram. You take 1200 joules/gram * (# of grams)/one mole [now you can cancel grams and it is now joules/mole.] Then convert the answer to kilojoules by dividing by 1000.
The heat of solution for barium chloride is -61.7 kJ/mol. To convert this to joules, multiply by 1000 to get -61,700 J/mol.
These are not compatible. The first is Energy per Temperature. The second is Energy per amount of matter.
To convert joules to kilojoules, divide by 1000. Therefore, 3923.7552 J is equal to 3.9238 kJ.
The first step is to calculate how many moles of Na are present in 1 mg (0.001 g) of Na. Since the molar mass of Na is 22.98977 g/mol, there are 0.000043 moles of Na. Next, convert the ionization energy to joules by multiplying by 1000 (1 kJ = 1000 J) to get 495,800 J/mol. Finally, multiply the number of moles by the energy per mole to find the total energy required: 495,800 J/mol * 0.000043 mol = 21.3 J.
DeltaG = DeltaH - TDeltaS dG = -54.32 kJ/mol - (54'32+273)K(-354.2J/molK) NB Thevtemperature is quoted in Kelvin(K) and the Entropy must be converted to kJ by dividing by '1000'/ Hence dG = - 54.32kJ/mol - (327.32K)(-0.3542 kJ/molK) NB The 'K' cancels out. Then maker the multiplication dG = -54/32 kJ/mol - - 115.94 kJ/mol Note the double minus; it becomes plus(+). Hence dG = -54.32kj/mol + 115.94 kJ/mol dG = (+)61.61 kJ/mol Since dG is positive, the reaction is NOT thermodynamically feasible.
Heat capacity is in the measurement of (kilo)Joules per mol degree Kelvin (J/mol K) Specific heat capacity is in joules/gram degree Kelvin (J/ gram K) Converting between the two is rather simple. To convert to specific heat capacity, divide the molar heat capacity by the molar mass of the molecule in question. eg. ( J/ mol K) / (grams/mol ) = J/ gram K, because mols will cancel.
12 J/3 moles = 4 J/mole. Thus, H of reaction in kJ/mole = 0.004 kJ/mole
The change in energy is -312KJ/mol. E=2.18*10^-18[ (1/2^2)-(1/9^2) ] =5.1884*10^-19 J/atom Divide by 1000 to convert to Kj =5.1884*10^-22KJ/atom 5.1884*10^-22KJ/atom x 6.022*10^23 (avo. #) atoms = -312KJ/Mol 1 Mol
Multiply by avagardoes number
The work done by the system can be calculated using the formula: Work = Force × Distance × cos(θ). Since work is given as -2.37 kJ, we convert this value to joules (1 kJ = 1000 J). So, the work done is -2.37 kJ × 1000 J/kJ = -2370 J. Therefore, the work done by the system when it absorbs 650 J of energy during a change is -2370 J + 650 J = -1720 J or -1.72 kJ.