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3 years ago

9

A local FM radio station broadcasts at a frequency of 99.0 MHz. Calculate the energy of the frequency at which it is broadcastin

g. energy in kj/photon
1MHz=10^6 sec-1

1 answer:

ycow [4]3 years ago

4 0

Answer:

Sorry i don't know the answer

.

.

buy if you don't mind then i want to ask something

<u>Whi</u><u>ch</u><u> class</u><u> </u><u>ques</u><u>tion</u><u> </u><u>is</u><u> </u><u>thi</u><u>s</u><u>?</u>

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The most appropriate answer is D ! temperature !

as when temperature increases KE increases and the collision factor and frequency increases !

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4 years ago

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Which of the following are found outside the nucleus?

lora16 [44]

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electrons
Reason:
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3 years ago

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At 25 °C, only 0.0990 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB3(s)↽−−

maksim [4K]

Answer:

$Ksp=2.59x10^{-3}$

Explanation:

Hello,

In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:

$AB_3\rightleftharpoons A^{3+}+3B^-$

The concentrations of the A and B ions in the solution are:

$[A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M$

$[B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M$

Then, as the solubility product is defined as:

$Ksp=[A][B]^3$

Due to the given dissociation, it turns out:

$Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}$

Regards.

4 0

4 years ago

Which of the following requires the most input of energy? A) Melting a substance B) Raising the temperature by 1 degrees c C) De

Butoxors [25]

Answer:

The correct answer is D Vaporizing a substance.

6 0

3 years ago

A zinc rod is placed in 0.1 MZnSO4 solution at 298 K. Write the electrode reaction and calculate the potential of the electrode.

yanalaym [24]

<u>Answer:</u> The potential of electrode is -0.79 V

<u>Explanation:</u>

When zinc is dipped in zinc sulfate solution, the electrode formed is $Zn^{2+}(aq.)/Zn(s)$

Reduction reaction follows: $Zn^{2+}(0.1M)+2e^-\rightarrow Zn(s);(E^o_{Zn^{2+}/Zn}=-0.76V)$

To calculate the potential of electrode, we use the equation given by Nernst equation:

$E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

$[Zn]=1M$ (concentration of pure solids are taken as 1)

$[Zn^{2+}]=0.1M$

Putting values in above equation, we get:

$E_{cell}=-0.76-\frac{0.059}{2}\times \log(\frac{1}{0.1})\\\\E_{cell}=-0.79V$

Hence, the potential of electrode is -0.79 V

4 0

3 years ago