80.0mL of a 1.00M HCl solution at 20.00Celsius is mixed with 80.0mL of a 1.00M Ca(OH)2 solution at 20.00 Celsius. If the final temperature of the mixture is 35.00*C, determine the heat of neutralization in kJ/mole. Assume a density of 1.00 g/cm3 and a specific heat of 4.184 J/goC. How do I approach this step by step?

these nuts

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally