Apan of liquid water freezes when you place it outside on a cold day. liquid water has greater molecular disorder than ice does. is the freezing process, then, an exception to the law of entropy?

No.

Explanation:

The freezing process is not an exception, simply because the pan of liquid water cannot be considered an isolated system.

In fact, the law of entropy holds for isolated systems only: the entropy of an isolated system can never decrease. However, in this case the pan of liquid water is not an isolated system, since it exchanges energy with the surrounding. In particular, it gives thermal energy away to the environment, so its temperature decreases and it freezes: the entropy of the pan of water decreases, but the total entropy of the system (pan of water + environment) does not decrease.

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂