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12345 [234]
2 years ago
6

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Chemistry
1 answer:
shusha [124]2 years ago
3 0

This problem is asking for the rate of disappearance of gaseous nitrogen, given the rate of appearance of ammonia and the chemical reaction. At the end, the result turns out to be -0.228 M/s.

<h3>Rates of appearance and disappearance</h3>

In chemical kinetics, one of the most relevant calculations are based on rates of appearance and disappearance of chemical species in a chemical reaction. This can be calculated via rate portions based on the stoichiometric coefficients in the reaction.

Thus, for this problem, one can write:

\frac{r_{N_2}}{-1} =\frac{r_{H_2}}{-3} =\frac{r_{NH_3}}{2}

Where the rate of appearance or disappearance is divided by the stoichiometric coefficient. Therefore, one can solve for the rate of disappearance of N2 with:

\frac{r_{N_2}}{-1} =\frac{r_{NH_3}}{2}\\\\r_{N_2}=\frac{r_{NH_3}*-1}{2}\\\\r_{N_2}=\frac{0.456M/s*-1}{2}\\\\r_{N_2}=-0.228M/s

Learn more about chemical kinetics: brainly.com/question/26351746

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A sheet of aluminum measures 55.0 cm by 23.3 cm. If the volume is 2.90 cm^3 what is the thickness of the foil?
tia_tia [17]

Answer:

0.0023 cm

Explanation:

The following data were obtained from the question:

Length (L) = 55 cm

Breadth (B) = 23.3 cm

Volume (V) = 2.9 cm³

Thickness (T) =?

Thus, we can obtain the thickness of the aluminum foil as illustrated below:

V = L × B × T

2.9 = 55 × 23.3 × T

2.9 = 1281.5 × T

Divide both side by 1281.5

T = 2.9/1281.5

T = 0.0023 cm

Therefore, the thickness of the aluminum foil is 0.0023 cm

3 0
3 years ago
5) When heated in a flame, the element Indium emits electromagnetic radiation with a distinctive indigo blue
grandymaker [24]

Answer:

\lambda=451nm

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by using the following equation, defined in terms of energy, Planck's constant, wavelength and speed of light:

E=\frac{hC}{\lambda }

Thus, we solve for the wavelength as shown below:

\lambda=\frac{hC}{ E}

And finally plug in the energy, Planck's constant and speed of light to obtain:

\lambda=\frac{6.6261 x 10^{-34} m^2 kg / s*3x10^8m/s}{4.405x10^{-19}m^2kg/s^2}\\\\\lambda=4.513x10^{-7}m*\frac{1nm}{10^{-9}m} \\\\\lambda=451nm

Regards!

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3 years ago
What aids in the digestion of fats and eliminates waste products?
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Explanation:

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Lostsunrise [7]

Answer:

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3 years ago
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