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Inessa [10]
3 years ago
11

12. A beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. Will it float in water,

Chemistry
1 answer:
Greeley [361]3 years ago
5 0

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:

density=\frac{mass}{volume}

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:

density=\frac{4900 g}{5400 mL}

Solving:

<u><em>density=0.907 </em></u>\frac{g}{mL}<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 \frac{g}{mL}, 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>

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When gaseous F2 and solid I2 are heated to high temperatures, the I2 sublimes and gaseous iodine heptafluoride forms. If 350. to
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P_{I2}=0.033atm

Explanation:

Hi, the first step is to calculate how much F2 there is in the container:

<u>Fluorine can be considered as an ideal gas</u> (given that is non-polar and has a small molecule). Using the ideal gas formula:

n=\frac{V*P}{T*R}

Where:

P=350 torr * \frac{1 atm}{760 torr}=0.46atm

T=250K

V=2.5 L

R=0.082 \frac{atm*L}{mol*K}

n=\frac{2.5L*0.46atm}{250K*0.082 \frac{atm*L}{mol*K}}

n=0.056 mol

Now, the mols of iodine:

n_{I2}=\frac{2.5g}{253.8 g/mol}

n_{I2}=9.85*10^{-3}mol

<u>The chemical reaction described is the following</u>:

I_2(g) + 7 F_2(g) \longrightarrow 2 IF_7(g)

In this case, the limitant reactant is the fluorine:

1) The 0.056 mol of F2 gives 8*10^{-3} mol of IF_7 and consumes 8*10^{-3} mol of I2.

2) At the end, in the conteiner we have:

8*10^{-3} mol of IF_7

0 mol of F_2

9.85*10^{-3}-8*10^{-3}=1.85*10^{-3} mol of I_2

In total: 9.85*10^{-3}mol .All in 2.5 L at 550 K

<u>The final pressure</u>:

P=\frac{T*R*n}{V}

P=\frac{550K*0.082 \frac{atm*L}{mol*K}*9.85*10^{-3}mol}{2.5L}

P=0.176 atm

The partial pressure:

P_{I2}=0.176atm*\frac{1.85*10^{-3} mol (I2)}{9.85*10^{-3} mol (total)}

P_{I2}=0.033atm

<em>Note: this partial pressure is calculated by the Dlaton's principle</em>

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