Given:
H soln = -33.3 KJ/mol
number of mol = 0.180 mol
use:
Q = H soln * number of mol
= (-33.3 KJ/mol)*0.180 mol
= -5.99 KJ
= -5990 J
So, heat released = 5990 J
Answer: 5990 J
14 Question ( point) Q See page 382 Adding nitric acid to water is quite exothermic. of HNO3 is-33.3 kJ mol, then how much heat is evolved by dissolving 0 180 mol HND If the ΔΗΡ οι in 1000m of water?...
6 Question (1point) See page 228 Adding nitric acid to water is quite exothermic. Calculate the temperature change (AT) when 100.0 mL of water (d=1.00 g/mL) at 20.0°C [c-75.3 J/(mol-°C)] and 10.0 mL of concentrated HNO(16.0 M,AHin-33.3 kJ/mol) are combined. 1st attempt See Periodic Table See Hint Assume the concentrated HNO3 has a density equal to that of water and its addition doesn't add a significant amount of water to the system. °C
Adding nitric acid to water is quite exothermic. Calculate the temperature change (AT) when 100.0 ml of water (* 100 g/muat 19.2°C [p = 75.3J/mol. °C)) and 10.0 mL of concentrated HNO3(16.5 M, AHson --33.3 kJ/mol) are combined. 1st attempt Il See Periodic Table See Hint Assume the concentrated HNO3 has a density equal to that of water and its addition doesn't add a significant amount of water to the system.