Can someone please answer before 1 pm . All data that is needed to answer the qu

Question

Can someone please answer before 1 pm . All data that is needed to answer the questions is in the first photo ! Record your data with appropriate labels and units. Part I: Mass of cube: Trial 1:30les Trial 2: rial 3: Heightif cuber Tral en Trial 2: a Cm Trial 3 lal widh of caber: Tial Tal0eCm Tril30lalom Part II: 3,2mL Volume of water in graduated cylinder: Volume of water and cube in graduated cylinder: Trial 1 Trial 2 Trial 3: - Trial 1: Trial 2: Trial 3: . Part III: Follow the example above and neatly and thoroughly record your data Mass of beaker Trenl 1:2r.oon | Trai 1 2.om-ai l 3.290 8.01% Mas of beakr + . Trail 1 :39. Coed Trail 2..39.7rail 3:39.10"; omL Sulution 31,7s

Explanation / Answer

part (i) :-

Mass of the cube = m

Volume of the cume = a3 = (lenth * height *width )

Density (P)= mass/ volume=m/v

Average density for Part (i) = Sum of all three densities /3 = 1.423 g/cm3

Part (ii):-

take mass of the cube from part (i) .

Volume of the cube for each trial = Volume of water and cube in the cylider - volume of water in the cylinder

Converting each volume in cm. (1ml = 1cm3)

Therefore , density calculation table :-

Average density for part (ii) = 0.51356 g/cm3

1.The trial 2 from Part (i) is the correct measurement as according to property of a cube , its Length , breadth and width should be equal

Trial 2 from part (i) only satisfies the specific criteria.

2. As the cube is dry, we know its exact volume and hence exact density. If say th cube had already started melting after volume calculations, then it will effect its volume and as density is inversely proportional to volume , it will increase as the volume decreases.

3. (I'm quite unsure of part 3 here) ,

Trial 1 Trial 2 Trial 3 P= (m/v) 1.3934 g/cm3 1.4768 g/cm3 1.3982 g/cm3

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