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Ira Lisetskai [31]
3 years ago
11

Balance:n² + h² ➡️ nh³ ​

Chemistry
1 answer:
seropon [69]3 years ago
5 0

Answer:

N₂ +  3H₂    →       2NH₃

Explanation:

Chemical equation:

N₂ +  H₂    →       NH₃

Balanced chemical equation:

N₂ +  3H₂    →       2NH₃

Step 1:

N₂ +  H₂    →       NH₃

left hand side:                       Right hand side

N =  2                                     N = 1

H = 2                                      H = 3

Step 2:

N₂ +  3H₂    →       NH₃

left hand side:                       Right hand side

N =  2                                     N = 1

H = 6                                      H = 3

Step 3:

N₂ +  3H₂    →       2NH₃

left hand side:                       Right hand side

N =  2                                     N = 2

H = 6                                      H = 6

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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
S-2-iodooctane lost his opticaly activity in soln on tratment wwithNaI. Explain
IceJOKER [234]

Answer:

Repeated SN2 reactions occur leading to the formation of a racemic mixture

Explanation:

S-2-iodooctane is a chiral alkyl halide with an asymmetric carbon atom. The presence of an asymmetric carbon atom implies that it can rotate plane polarized light and thus lead to optical isomerism. The two configurations of the compound are R/S according to the Cahn-Prelong-Ingold system.

However, when S-2-iodooctane is treated with sodium iodide in acetone, repeated SN2 reactions occur since the iodide ion is both a good nucleophile and a good leaving group. Hence a racemic modification is formed in the system with time hence we end up with (±)- Iodooctane.

5 0
3 years ago
Please help (30 points)
aleksandrvk [35]

1) is chemical Bonds

3) Conservation of mass

5) compound

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7 0
3 years ago
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What are the charges on barium and fluorine in barium fluoride? 1 point Captionless Image Barium is +2 and fluorine is -2. Bariu
user100 [1]

Answer: Barium is +2, Fluorine is -1

Explanation:

The charge of barium is +2, and the charge of fluorine is -1. You can determine this from the periodic table groups.

The formula for barium fluoride is thus BaF2.

8 0
3 years ago
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kolezko [41]
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