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2 years ago

A new radio station, KHEM-AM, broadcasts

Chemistry

1 answer:

Serga [27]2 years ago

7 0

Answer:

wavelength = 345.62 m

Explanation:

Given that,

Frequency of signal, f = 868 kHz

We need to find the wavelength of its signal.

The relation between wavelength and frequency is given by :

$c=f\lambda$

c is speed of light

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{868\times 10^3}\\\\\lambda=345.62\ m$

So, the wavelength of its signal is 345.62 m

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a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

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MM: 17.03 32.00 30.01

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Mass/g: 1.5 1.85

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3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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