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Maksim231197 [3]
2 years ago
11

3. Mothballs in a clothes closet gradually disappear over time. What happens to this material?

Chemistry
1 answer:
Svet_ta [14]2 years ago
4 0

Mothballs in a clothes closet gradually disappear over time, because their components sublime.

Mothballs are small balls of chemical pesticide and deodorant, sometimes used in closets where clothes are susceptible to damage from mold or moths. Mothballs are usually made of naphthalene, 1,4-dichlorobenzene or camphor. What all the compounds have in common is that they sublime at room temperature. Sublimation is a physical change in which matter changes from the solid to the gas state. Once the components are in the gaseous state, the diffuse, and that's why there are no visible remainings over time.

Mothballs in a clothes closet gradually disappear over time, because their components sublime.

You can learn more about sublimation here:

brainly.com/question/10936758

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See explanation.

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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

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