You have two beakers, one filled to the 100-ml mark with sugar (the sugar has a mass of 180.0 g) and the other filled to the 100-ml mark with water (the water has a mass of 100.0 g). you pour all the sugar and all the water together in a bigger beaker and stir until the sugar is completely dissolved.
b. which of the following is true about the volume of the solution? explain.
i. it is much greater than 200.0 ml.
ii. it is somewhat greater than 200.0 ml.
iii. it is exactly 200.0 ml.
iv. it is somewhat less than 200.0 ml.
v. it is much less than 200.0 ml.

V. It is much less than 200 ml.

Explanation:

The final volume of the sugar-water mixture is gonna be something very close to the volume of water itself. The reason to this contraintuitive answer is that sugar molecules can dissolve and "find spaces inside the water molecular structure". In other words in a sugar beaker or cup the volume is mostly free air, because its a crystalline net structure.

The big change is going to be in the density of the solution and the mass is going always to be preserved. So there will be

180 g from sugar + 100g from water = 280 g of total volume

Density=mass/volume

Density of water=100/100=1

Density of sugar=180/100=1.8

Density of solution aprox=280/100=2.8

  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

god i hope this !