
How do you convert from #"kJ/g"# to #"kJ/mol"# - Socratic
2018年3月10日 · Use the molar mass of the compound. Your goal here is to convert kilojoules per gram, "kJ g"^(-1), to kilojoules per mole, "kJ mol"^(-1), which is equivalent to saying that you need to convert the amount of energy per unit of mass to the amount of energy per unit of mole. In order to do that, you essentially need to use a conversion factor that will take you from grams to moles. As you know ...
Hess' Law - Chemistry - Socratic
Hess' Law describes the conservation of energy. That is, that regardless of the path taken during a chemical reaction or whether the chemical reaction was completed in one step or several, the enthalpy change in the reaction remains the same.
Calorimetry - Chemistry - Socratic
The specific heat capacity of liquid water is 4.18 kJ/g C, how would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C? A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K.
Convert 87.69 kJ/g into k/mol? + Example - Socratic
2018年3月14日 · CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(l) +Delta ...This site quotes an enthalpy of combustion for methane of -890*kJ*mol^-1, and when we quote this, we MEAN per mole of reaction as written...i.e. the combustion of 16*g of methane with stoichiometric dioxygen RELEASES 890*kJ...
How do I calculate Energy (kcal/g) if I know the Chemical ... - Socratic
2018年5月7日 · Now of course, since we know the molar masses of products and reactants, I could convert that #kJ*mol^-1# into #kJ*g^-1# of fuel, or if I am a real masochist into #kcal*mol^-1#, or #kcal*g^-1#. So the take home message is that #kJ*mol^-1# implies a #"MOLE QUANTITY"# of a specific reaction of reference. You have to be sufficiently flexible to ...
Given DeltaH_"combustion"^@ for butane is -2657*kJ*mol^-1
2017年3月22日 · "Heat of combustion of butane" -= 45.7xx10^3*J*g^-1 We have data for the stoichiometric equation: C_4H_10(g)+ 13/2O_2(g)rarr 4CO_2 + 5H_2O DeltaH_"rxn"=-2657*kJ*mol^-1 Note that I halved the equation because it makes the arithmetic a little bit easier.
The latent heat of vaporization of water is 2260 J/g. How many ...
2015年12月4日 · "2.26 kJ/g" For a given substance, the latent heat of vaporization tells you how much energy is needed to allow for one mole of that substance to go from liquid to gas at its boiling point, i.e. undergo a phase change. In your case, the latent heat of vaporization for water is given to you in Joules per gram, which is an alternative to the more common kilojoules per mole. So, you need to ...
The combustion of 20 g of a fuel in a calorimeter increases the ...
2017年12月17日 · The combustion of 20 g of a fuel in a calorimeter increases the temperature of 200 g of water by 5 °C. What is the enthalpy of combustion per gram of fuel? The specific heat capacity of water is #"4.184 J·°C"^"-1""g"^"-1"# .
"1.659 g" of sample benzoic acid was placed in a bomb ... - Socratic
2015年8月4日 · Basically, the exothermic DeltaH_C is the negatively-signed result of the following: The heat capacity of the calorimeter in "J/K" multiplied by the change in temperature in "K", the quantity plus RT_"room"Deltan_"gas", with Deltan_"gas" gotten from multiplying (1.659g)/(122.12g) by the change in moles of gas of: 2C_6H_5COOH(s) + 13O_2(g) -> 12CO_2(g) + 6H_2O(g) which is 5 "mol" *(1.659g)/(2 ...
The specific heat capacity of liquid water is 4.18 kJ/g C ... - Socratic
2015年10月26日 · "239 J" First thing first, you mistyped the specific heat of water, which should be c_"water" = 4.18"J"/("g" ""^@"C") Now, a substance's specific heat tells you how much heat is required to increase the temperature of "1 g" of that substance by 1^@"C". In the case of water, you would need "4.18 J" to increase the temperature of "1 g" of water by 1^@"C". Notice that your sample of water has a ...