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S_A_V [24]
3 years ago
15

What is the frequency, in hertz, of an electromagnetic wave with a wavelength of 625 nm?

Chemistry
1 answer:
Reika [66]3 years ago
6 0

Answer:- Frequency is 4.80*10^1^4Hz .

Solution:- The wavelength of electromagnetic radiation is given as 625 nm and we are asked to calculate the frequency. Frequency is inversely proportional to the wavelength and the formula used to calculate frequency when wavelength is given is:

\nu =\frac{c}{\lambda }

where, \nu is frequency, c is speed of light and \lambda is the wavelength.

Value of speed of light is \frac{3.0*10^8m}{s} . We need to convert the given wavelength from nm to m.

1nm=10^-^9m

So, 625nm(\frac{10^-^9m}{1nm})

= 6.25*10^-^7m

Let's plug in the values in the formula and calculate the frequency:

\nu =\frac{3.0*10^8m.s^-^1}{6.25*10^-^7m}

= 4.80*10^1^4s^-^1 or Hz.

So, the frequency of the electromagnetic wave is 4.80*10^1^4Hz .

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The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the
Mademuasel [1]

Answer:

(a) rate =  4.82 x 10⁻³s⁻¹  [N2O5]

(b) rate =   1.16 x 10⁻⁴  M/s

(c) rate =   2.32 x 10⁻⁴ M/s

(d) rate =   5.80 x 10⁻⁵ M/s

Explanation:  

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration   of   N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹  x 0.0240 M =  1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of  N₂O₅ is double the rate will double too:  2 x   1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 =  5.80 x 10⁻⁵ M/s

3 0
3 years ago
Read 2 more answers
hi, help me please with both number 4 and 5, please be sure of your answer, first to answer gets brainliest!​
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Answer:

4. mutualism

5. commensalism

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8 0
2 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

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