2.5 g 1 mol/18.02 g (-285.83) kJ/mol
The ∆Hfusion for water is 333.55 J/g. Thus, q = 98.5 g x 333.55 J/g = 32,855 J = 32.9 kJ (3 sig figs)
specific heat of ice = 2.09 J/g/degree2.09 J/g/deg x 800 g x 5 deg = 8360 J = 8.4 kJ∆Hfusion = 334 J/g0.8kg x 334 J/g x 1000 g/kg x 1 kJ/1000 J = 267 kJ8.4 kJ + 267 kJ = 275.4 kJ
This value is 73,12 kJ.
1.3 g / 58.123 g C4H10 x -147.6 kJ =
Melting of 183,3 g ice need 61,14 kJ.
298-299 kJ
-18 kj/mol
You have 42.5 g of water. 42.5g H20 / 18.02 g H2O (2.358 moles H2O)*(6.02 Kj/1mole H2O) = 14.1981 Kj =14.2 kJ
G=18 kJ/mol
Other comment is wrong 2.5 g 1 mol/18.02 g (-285.83) kJ/mol
∆G = ∆H - T∆S∆G = 27 kJ/mole - (500 deg)(0.09 kJ/mol-deg) ∆G = 27 kJ/mole - 45 kJ/mole ∆G = - 18 kJ/mole (Note the minus sign indicating the process is spontaneous)
4.3 g * 1 mol/18.02 g * 6.03 kJ/mol
4.3 g * 1 mol/18.02 g * 6.03 kJ/mol
2.5 g 1 mol/18.02 g (-285.83) kJ/mol
G = 18 kJ/mol
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ/mol