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NNADVOKAT [17]
3 years ago
5

Why are gamma rays not able to reflect off a mirrors surface

Chemistry
2 answers:
ElenaW [278]3 years ago
7 0

Answer:

the Ray's go right through it

Explanation:

the Ray's are so small they punch right through them

givi [52]3 years ago
7 0

Answer:

Explanation:

The reason why is based in something called the plasma frequency of the metal of a mirror. A metal, as you may know, is composed of a series of atom (ion, effectively) cores - nuclei, together with some, but not all, of their bound electrons - which contribute the remaining outermost electrons of their unbound forms to a communally shared "electron sea" - kind of like a giant, distributed omnidirectional covalent bond that extends all throughout the whole metal crystal (here we're just considering a single crystal for simplicity). The electrons are quantumed out all over the full extent of the crystal and effectively form a sort of "gas" throughout and permeating the metal.

When an electromagnetic wave approaches that gas, the free charges within it - the electrons - start oscillating, and as they do so, they set up another wave going outward at the same time as the first is going in. This begins as soon as the first wave begins to impinge.

However, if the wave oscillation is fast enough, the electrons can't keep up due to their mass, and thus they are unable to form the reflected wave. The frequency at which this occurs is called the metal's plasma frequency (and is inversely proportional to the square root of the mass, so that a high mass particle would have a lower plasma frequency). The name comes from the fact that the metal can be thought of in a sense as a kind of "solid plasma" - ions with free electrons, the difference with what most people think of as a "plasma" being here the ions are not free to move about of their own accord.

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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
ale4655 [162]

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

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Answer:

Explanation:

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