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Leni [432]
3 years ago
15

A photon of green light strikes an unknown metal and an electron is emitted. The voltage is set to 2 volts. The electron cannot

make the journey to the second plate. What can be said about a similar experiment done with violet light? A. An electron may or may not be emitted in the second experiment. It cannot be determined. B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate. C. An electron will be emitted in the second experiment, and it will make it to the second plate. D. An electron will not be emitted in the second experiment.
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
7 0

Answer:

The correct answer is:

An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate.

Explanation:

In fact, violet has higher frequency than green light. This means that photons on violet carry more energy than photons of green light (remember that the energy of a photon is proportional to it's frequency:

e = hf

, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted with also violet light, given the more energy transferred.

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5.82 760.<br> Х<br> 425.976<br> answer should have<br> sig.figs.<br> calculated answer:
Mekhanik [1.2K]

Answer:

2482.4177376

Explanation:

1: 5.82 760  times   425.976

2: You will find the answer that is 2482.4177376

<em><u>Hope this helps.</u></em>

7 0
3 years ago
If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
3 years ago
How many electrons in an atom with the atomic number of 45
Sergio039 [100]

Answer:

Number of Protons: 45

Number of Neutrons: 58

Number of Electrons: 45

Explanation:

Have a nice day

4 0
3 years ago
Explain how you were able to use your knowledge of how different types of blood react with anti A, anti B, and anti Rh antibodie
dybincka [34]

Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.

Explanation:

Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.

A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.

A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.

B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.

B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.

Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.

Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.

3 0
3 years ago
Hello! Id like if someone could tell me this answer, thank you so much!
DaniilM [7]

Answer:

Option D. ZnCl₂ and H₂

Explanation:

From the question given above, the following equation was obtained:

2HCl + Zn —> ZnCl₂ + H₂

Products =?

In a chemical equation, reactants are located on the left side while products are located on the right side i.e

Reactants —> Products

Now, considering the equation from the question i.e

2HCl + Zn —> ZnCl₂ + H₂

The products are ZnCl₂ and H₂ because from our discussion above, we said that products are only located on the right side of chemical equation.

Thus, option D gives the correct answer to the question.

4 0
3 years ago
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