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Blizzard [7]
2 years ago
9

Which observation provided Albert Einstein the clue that he needed to explain the photoelectric effect? Light is made up of extr

emely small particles. Particles that are emitted from a strip of metal are electrons. Shining light on a metal strip produces emissions. Energy of electrons depends on light’s frequency, not intensity.
Chemistry
2 answers:
Aliun [14]2 years ago
7 0

The correct answer is

Energy of electrons depends on light’s frequency, not intensity.

As per photoelectric effect, if we incident a light on metal surface it will results into emission of electron from it

if we increase the number of photons the number of electrons will increase however if we increase the frequency the number of photons will not increase

While if we increase frequency the energy of electrons will increase as

Energy of photon = Work function of metal + kinetic energy of electrons

Mice21 [21]2 years ago
4 0
Energy of electrons depends on light's frequency not intensity is the correct option.

Hope this helps, have a great day/night ahead!
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Condensation is the change of the physical state of matter from the gas phase into the liquid phase.

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Answer:

\boxed{\text{2408 min}}

Explanation:

The integrated rate law for radioactive decay is

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1. Calculate the decay constant

\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}\\\\

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}

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At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
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At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

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By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

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4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

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