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 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.

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.

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...

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 ...

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 ...

2017年5月17日 · #2.4 * 10^(-4) color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(39.997color(red)(cancel(color(black)("g")))) = 6.00 * 10^(-6)# #"moles NaOH"# You know that the enthalpy of dissolution when #6.00 * 10^(-6)# moles of sodium hydroxide are dissolved in water, so use this info to find the enthalpy of dissolution when #1# mole of the salt dissolves

2015年3月31日 · ΔH = "-279 kJ" This is a Hess's Law problem. Our target equation is 2C(graphite) + 2H₂(g) + ½O₂(g) → C₂H₅OH(l); ΔH = "?" We have the following information: 1. C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O (l); ΔH= "-1367 kJ" 2. H₂(g) + ½O₂(g) → H₂O(l); ΔH = "-286 kJ" 3. C(graphite) + O₂(g) → CO₂(g); ΔH = "-394 …

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 ...

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"# .