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3 years ago

ASAP WILL GIVE BRAINLIEST!!

Chemistry

1 answer:

Nataly_w [17]3 years ago

6 0

Answer:

The answer to your question is a) N₂ b) 3.04 g of NH₃

Explanation:

Data

mass of H₂ = 2.5 g

mass of N₂ = 2.5 g

molar mass H₂ = 2.02 g

molar mass of N₂ = 28.02 g

molar mass of NH₃ = 17.04 g

Balanced chemical reaction

3H₂ + 1 N₂ ⇒ 2NH₃

Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22

Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1

Conclusion

The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.

28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃

2.5 g of N₂ -------------------- x

x = (2.5 x 2 x 17.04) / 28.02

x = 85.2 / 28.02

x = 3.04 g of NH₃

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3 years ago

What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K

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Answer: The de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

Explanation:

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

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where,

E = kinetic energy of the particles = ?

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T = temperature of the particle = 30 K

Putting values in above equation, we get:

$E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J$

Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used: 1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or $2\times 10^{-3}kg$

According to mole concept:

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

As, $6.022\times 10^{23}$ number of hydrogen molecules has a mass of $2\times 10^{-3}kg$

So, 1 molecule of hydrogen will have a mass of = $\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg$

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of 1 hydrogen molecule = $3.32\times 10^{-27}kg$

$E_k$ = kinetic energy of the particle = $6.21\times 10^{-22}J$

Putting values in above equation, we get:

$\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}$

$\lambda=3.26\times 10^{-10}m=3.26\AA$ (Conversion factor: $1\AA=10^{-10}m$ )

Hence, the de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

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Keith_Richards [23]

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3 years ago

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3 years ago