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igor_vitrenko [27]
3 years ago
15

A solid rubber ball with a mass of 5.9 grams sank in water. A hollow rubber ball with the same mass floats in water. Explain why

this might be. Please answer ASAP!
Chemistry
1 answer:
Stells [14]3 years ago
4 0

Answer:

The hollow ball is more buoyant than the solid ball.

Explanation:

The hollow ball will float while the solid ball sinks because the volume of water displaced by the hollow ball is more than that displaced by the solid ball.

An object floats when it displaces a volume of water equal to its weight.  A solid ball is more compact, and hence, does not displace a lot of water when dropped into a bowl full of water. On the other hand, the hollow ball will have a larger curved surface area, since it has no fill. This will make it displace a larger volume of water and hence make it float.

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
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Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

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Explanation:

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