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Aleksandr-060686 [28]
3 years ago
7

Telephone signals are often transmitted over long distances by microwaves. what is the frequency of microwave iradiation with a

wavelength of 3.50 cm
Chemistry
2 answers:
masha68 [24]3 years ago
7 0
The computation for this problem would be:

The formula that we will be using is: f = c/λ

Where:

c = speed of light = 3.00 x 10^5 km/s convert it to meters, so: 3.00 x 10^8 m/s 

λ = wavelength of the microwave radiation = 3.50 cm convert this to meters, so: 0.035 m 

f = frequency (in Hertz) 
f = c/λ

= 3.00 x 10^8 m/s / 0.035 m 

f = 8.57 x 10^9 Hz Frequency 
Alborosie3 years ago
6 0
<span>c = speed of light = 3.00 x 10^5 km/s = 3.00 x 10^8 m/s
  λ = wavelength of the microwave radiation = 3.50 cm = 0.035 m
  f = frequency (in Hertz) = to be determined
    f = c/λ = 3.00 x 10^8 m/s / 0.035 m
  f = 8.57 x 10^9 Hz Frequency</span>
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An exciton is: A charged particle made of electrons A charged particle made of holes A neutral particle made of an electron and
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3 years ago
What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen
Blizzard [7]

Answer: Mg_{1}Cl_{2}O_{8}

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg=\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles

Moles of Cl = \frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles

Moles of O =\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = \frac {0.45}{0.45}=1

For Cl = \frac {0.89}{0.45}=2

For O= \frac {3.58}{0.45}=8

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is Mg_{1}Cl_{2}O_{8}

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3 years ago
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