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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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When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

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Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

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Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

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Please see attachment

Explanation:

Please see attachment

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