Find the enthalpy of neutralization of hcl and naoh. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 mol dm-3 naoh. the temperature rose from 298 k to 317.4 k. the specific heat capacity is the same as water, 4.18 j/k g. a. -101.37 kj b. 7055 kj c. 10,1365 kj

The correct answer is A : -101.37 KJ

Explanation:

Specific heat, s = 4.18 J/Kg

Density of water is 1g/cm3 , so 174 cm3 of total solution is = 174 g

Total mass of reaction mixture is = 174 g

Rise in tempetrature, ΔT =  317.4 K -298 K = 19.4 K

87 cm3 of 1.6 mol dm-3 of HCl = 87 cm3 of 1.6 mol dm-3 NaOH

1.6 M solution means that 1000 cm3 of solution has 1.6 moles.

So, 87 cm3 of 1.6 M solutions = 0.1392 M of HCl and NaOH

Heat evolved (q) = m x s x  ΔT

                   = 174 g x 4.18 J/Kg  X 19.4 K

                   =14110.0 J = 14.110 KJ (for exothermirmic reaction -14.110 KJ )

 Enthalpy of neutralization = -14.110 KJ/ 0.1392 = - 101.37 KJ

it is called acid bonding

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.