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3 years ago

We have a photon with the wavelength of 605 nm. What is the frequency of this photon?

Chemistry

2 answers:

tester [92]3 years ago

6 0

Answer:

ν=4.96x10^14 Hz

Explanation:

c=λν

λ=wavelength

ν=frequency

in this problem:

c=3x10^8

λ=605x10^-9

so ν=c/λ

v=(3x10^8)/(605x10^-9)

v=4.96x10^14

please give thanks :)

Kipish [7]3 years ago

3 0

Answer:

Explanation:

Speed of light = Wavelength x Frequency.

605 nm x Frequency = 3x10^8 m/s

6.05 x 10^-7 m x Frequency = 3x10^8 m/s

Frequency = 4.96 x 10^14 Hz

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The order of the reaction is 1.

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The rate law of the reaction ;

$R=k[N_2O_5]^x$

The rate of the reaction from T = 0 s to 25 s, $[N_2O_5]=1.000 M$ to $[N_2O_5]=0.822 M$ respectively.

$R=-\frac{0.822 M-1.000 M}{25s-0 s}=0.00712 M/s$

$0.00712 M/s=k[0.822 M]^x$ ..[1]

The rate of the reaction from T = 25 s to 50 s, $[N_2O_5]=0.822 M$ to $[N_2O_5]=0.677 M$ respectively.

$R=-\frac{0.677 M-0.822 M}{50s-25 s}=0.0058 M/s$

$0.00646 M/s=k[0.677 M]^x$ ..[2]

[1] ÷ [2]

$\frac{0.00712 M/s}{0.0058 M/s}=\frac{k[0.822 M]^x}{k[0.677 M]^x}$

Solving for x:

x = 1.05 ≈ 1

The rate law of the reaction ;

$R=k[N_2O_5]^1$

The order of the reaction is 1.

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4 years ago