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

lithosphere and asthenosphere

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm