An oil bath maintained at 50.5°c loses heat to its surroundings at the rate of 4.68 kj/min. its temperature is maintained by an electrically heated coil with a resistance of 60 operated from a 110 v line. a thermoregulator switches the current on and off. what fraction of the time will the current be turned on?

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained,

Heat loss at rate, q = 4.68 kJ/min

Resistance, R =

Operating Voltage,

Now,

Power that the resistor releases,

The fraction of time for the current to be turned on:

t = 0.3852

  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