Agraduated cylinder if filled with water to a level of 40.0 ml. when a piece of paper of copper is lowered into the cylinder , the water level rises to 63.4ml. find the volume of the copper sample. if the density of the copper is 8.9g/cm³, what is the mass

Given:

Water level is 40mL

Water level rises to 63.4mL after copper is placed

Density is 8.9g/cm3

Required:

Mass of the object

Solution:

The displaced water level is the volume of the copper

63.4mL – 40mL = 23.4mL

Since 1mL = 1cm3, therefore 23.4mL is 23.4cm3

The density formula is D = M/V where D is density, M is mass of the object and V is the volume of the object. Rearranging we get M = DV.

M = DV

M = (8.9g/cm3)(23.4cm3)

M = 208 g

  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

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