How many molecules are there in 985 mL of nitrogen at 0.0 degrees C and 1.00x10^-6 mm Hg? (It would be better if work or steps to solve problem are given but if not its fine).

3.48 x 1013 N2 molecules

step-by-step explanation:

Lets set up our equation first

P = 1.00 x 10-6 mm Hg T = 0.0° C + 273 = 273 K

We are given the V = 985 mL ,

R = 0.0821 L·atm/mol·K

Now use the ldeal gas law, but we are solving n, amount of substanvce

PV = nRT, we will change this equation to ;

n = PV/RT

n = 1.00 x 10-6 mm x 1 atm/760 mm x 985 mL x 1 L/103

mL/

(0.0821 L·atm/mol·K x 273 K) = 5.78 x 10-11 moles N2

nmolecules = 5.78 x 10-11 moles N2 x 6.02 x 1023 N2 molecules/1 mol N2

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.