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Taking into account the definition of density and Archimedes' principle, the beginner bowling ball will float on the water.
But first it is neccesary to know that density is a quantity referred to the amount of mass in a certain volume of a substance or a solid object.
In other words, the density is the relationship between the weight (mass) of a substance and the volume that the same substance occupies.
The expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:
In this case, a beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. This is:
- mass= 4.9 kg= 4900 g (being 1 kg= 1000 kg)
- volume= 5.4 L= 5400 mL (being 1L=1000 mL)
Replacing in the definition of density:
Solving:
<u><em>density=0.907 </em></u><u><em /></u>
On the other hand, Archimedes' principle says that an object immersed in a liquid experiences an upward vertical force equal to the weight of the volume of the dislodged liquid.
The sinking or floating of an object is determined by its density with respect to that of the liquid in which it is submerged.
Considering water as the liquid where the object is submerged in this case, an object with a higher density than water will sink. In contrast, an object with a lower density than water will float.
In this case, considering that water has a density of 1 , the bowling ball for beginners has a lower density. This indicates that, having a lower density than water, the object will float.
In summary, the beginner bowling ball will float on the water.
Learn more about density:
- <u>brainly.com/question/952755?referrer=searchResults </u>
- <u>brainly.com/question/1462554?referrer=searchResults</u>
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to = 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is = 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³