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 can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.