How many grams of water can be frozen at 0 ^oC with the removal of -129.8 J? (Specific Heat = 4.2 𝐽/𝑔^o𝐶, Enthalpy of fusion 6010 J/mol, Enthalpy of vaporization = 40700 J/mol)

1 yd2 = 8.361×10-25 Tm24.14×1025 yd2 = 4.4×1025 × 8.361×10-25 Tm2= 36.7884 Tm2

d,e

pH= -log[H+]pH= -log[1×10^-3]pH= 32) pH= -log[3.6×10^-5]pH= 5-log3.6pH= 4.44

H= mcΔtm= H/cΔt= 42000/(4.2×100) = 100g

Freezing point of water = 0°CBoiling point of water = 100°CCwater = 4.2 Jg-1°C-1q = m × c × ∆T m = q/∆T × c= 42.6/(100-0) × 4.2= 0.01g

a) 4x3

D) 2.408 x 1024 molecules of water

147N+42He→178O+11H714N+24He→817O+11H is an example of a fission reaction

Energy (n), angular momentum (ℓ), magnetic moment (mℓ), and spin (ms).

1 cubic metre =1000 litres10l = 1×10/1000 = 0.01B. 0.01m3