HELP ASAP - How much energy is lost when 80.0 grams of steam at 100.C turns into liquid water at 100.C?

Q = 43,000 calories = 180,000 joules = 180 Kilojoules

Explanation:

Heat flow for phase change can be defined as Q = m·ΔHv  

ΔHv = Heat of Vaporization for water = 540 calories/gram = 2259 joules/gram = 2.259 Kilojoules/gram

m = mass of object of interest ( in this problem 80 grams steam 100°C converting to 80 grams of water at 100°C

Q = m·ΔHv = 80g x 540 cals/g =43.200 calories x 4.184 joules/cal = 180,748.8 joules (calculator answer) => 180,000 joules (2 sig. figs.) = 180 Kj

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :