An object will float at the surface of a liquid if the mass of the object is less than the mess of the liquid that it displaces. a spherical vessel (diameter = 5cm) when empty has a mass of 12g. what is the greatest volume of water that can be placed in the vessel and still have the vessel float at the surface of benzene?

45.44 mL

Explanation:

Given:

Assume:

According to the given question, an object floats at the surface of a liquid if the mass of the object is less than the mass of the liquid displaced by it.

Let the greatest volume of water that can be put in the vessel so that it still floats at the surface of the benzene. Then the sum of the masses of the volume of water in the vessel and the empty vessel is equal to the mass of the volume of benzene displaced by the whole vessel.

i.e., Mass of water in the vessel + Mass of the empty vessel = Mass of the benzene displaced by the whole vessel

Hence, the greatest volume of water that can be placed in the vessel such that it still floats on the benzene surface is 45.44 mL.

  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