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Arturiano [62]
3 years ago
14

What is the de Broglie wavelength, in cm, of a 11.0-g hummingbird flying at 1.20 x 10^2 mph?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:1.123 x 10^-31cm

Explanation:

mass of humming bird=  11.0g

speed= 1.20x10^2mph

but I mile = 1.6m

1km=1000

I mile = 1.6x10^3m

1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2

=1.932 x10^5m

recall that  

1 hr= 60 min

1 min=60 secs, 1hr=3600s

Speed = distance/ time

=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s

m= a 11.0g= 11.0 x 10^-3kg

h=6.626*10^-34 (kg*m^2)/s

Wavelength = h/mu

= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)

6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m

but 1m = 100cm

1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm

de broglie wavelength of humming bird = 1.123 x 10 ^-31cm

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<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of CO_{2}

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<h2>Why?</h2>

We are asked to calculate the number of moles of the given gas, also, we  are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

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P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

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R=62.363\frac{mmHg.L}{mol.K}

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

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PV=nRT

n=\frac{PV}{RT}

n=\frac{PV}{RT}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole

Hence, we have that there were produced 0.120 moles of CO_{2}

n=0.120mol

Have a nice day!

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3 years ago
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