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3 years ago

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A sound wave traveling through a certain freshwater lake has a frequency of 349.2 Hz and a wavelength of 4.25 m. What is the spe

ed of a sound wave in this water? (Round your answer to three significant figures.)

2 answers:

erica [24]3 years ago

7 0

Answer:

ν = 1483.939 m/s

Explanation:

Wave propagation speed (ν):

ν = λ / T

∴ Wavelength ( λ ) = 4.25 m

∴ Period ( T ) = 1 / f

∴ frequency ( f ) = 349.2 Hz ( s-1 )

⇒ T = 1 / 349.2 s-1

⇒ T = 2.864 E-3 s

⇒ ν = (4.25 m) / (2.864 E-3 s)

⇒ ν = 1483.939 m/s

kramer3 years ago

6 0

<u>Answer:</u> The wave speed of the given sound wave is $1.48\times 10^3m/s$

<u>Explanation:</u>

Wave speed is defined as the product of its wavelength and frequency.

Mathematically,

$\text{Wave speed}=\lambda\times \nu$

where,

$\lambda$ = wavelength of the wave = 4.25 m

$\nu$ = frequency of the wave = 349.2 Hz = $349.2s^{-1}$

Putting values in above equation, we get:

$\text{Wave speed}=4.25m\times 349.2s^{-1}\\\\\text{Wave speed}=1484.1m/s=1.48\times 10^3m/s$

Hence, the wave speed of the given sound wave is $1.48\times 10^3m/s$

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When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

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<u>Answer:</u>

<u>For a:</u> The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

<u>For b:</u> The mass of nitric acid formed is 54.81 grams

<u>For c:</u> The mass of nitric acid formed is 206 grams

<u>Explanation:</u>

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

<u>For a:</u>

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

<u>For b:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

<u>For c:</u>

<u>For nitrogen dioxide:</u>

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

<u>For water:</u>

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

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