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kvasek [131]
3 years ago
15

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga

seous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that liters per second of dinitrogen are consumed when the reaction is run at and. Calculate the rate at which ammonia is being produced.
Chemistry
1 answer:
VashaNatasha [74]3 years ago
6 0

The question is incomplete, complete question is :

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 348 liters per second of dinitrogen are consumed when the reaction is run at 205°C and 0.72 atm. Calculate the rate at which ammonia is being produced.

Answer:

The rate of production of ammonia is 217.08 grams per second.

Explanation:

N_2+3H_2\rightarrow 2NH_3

Volume of dinitrogen used in a second = 348 L

Temperature of the gas = T = 205°C = 205+273 K = 478 K

Pressure of the gas = P = 0.72 atm

Moles of dinitrogen = n

n=\frac{PV}{RT}=\frac{0.72 atm\times 348 L}{0.0821 atm L/mol K\times 478 K}=6.385 mol

According to reaction, 1 mole of dinitriogen gives 2 mole of ammonia.Then 6.385 moles of dinitrogen will give:

\frac{2}{1}\times 6.385 mol=12.769 mol

Mass of 12.769 moles of ammonia;

12.769 mol 17 g/mol = 217.08 g

217.08 grams of ammonia is produced per second.So, the rate of production of ammonia is 217.08 grams per second.

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Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a
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The complete question is shown on the first uploaded image

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a) Splitting pattern for Ha= 2+1 , Triplet

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Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

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Splitting pattern for Hd= 0+1 , Singlet

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Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image  

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

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Considering the third question

The name of the given molecule  is gotten according to longest carbon chain  = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

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