(1.6): What volume of oxygen, at 18°C and 750 torr, can be obtained from heating 100gm of potassium chlorate, KCIO,?

Volume of oxygen obtained = 30 L

Explanation:

Given data:

Mass of KClO₃ = 100 g

Pressure  = 750 torr = 750/760 = 0.98 atm

Temperature  = 18°C = 18+273 = 291 K

Volume of oxygen obtained = ?

Solution:

Chemical equation:

2KClO₃    →     3O₂ + 2KCl

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles = 100 g/ 122.55 g/mol

Number of moles = 0.82 mol

Now we will compare the moles of KClO₃ with oxygen.

              KClO₃            :               O₂

                  2                :                 3

                0.82            :              3/2×0.82 = 1.23 mol

Volume of oxygen:

PV = nRT

0.98 atm × V = 1.23 mol ×0.0821 atm.L/mol.K × 291 K

0.98 atm × V = 29.4 atm.L

V = 29.4 atm.L /0.98 atm

V = 30 L

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.