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3 years ago

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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Answer:

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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

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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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