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myrzilka [38]
3 years ago
5

How many molecules (not moles) of nh3 are produced from 6.89×10−4 g of h2?

Chemistry
1 answer:
AfilCa [17]3 years ago
7 0
First, let's find the moles of hydrogen used to produce the ammonia
      mol = mass ÷ molar mass
if the mass of hydrogen = 6.89 × 10⁻⁴ g
& the molar mass of hydrogen = 2 g/mol
      then moles of hydrogen = (6.89 × 10⁻⁴ g) ÷ 2 g/mol
                           moles of H₂ = 3.445 × 10⁻⁴  mol

Next, balanced eq'n for the production of ammonia from hydrogen gas and nitrogen gas:
                N₂   +   3H₂   →  2NH₃

Now, the mole ratio (ratio of the coefficients used to balance the equation) of H₂ : NH₃ is  3 : 2

∴ if the moles of H₂ = 3.445 × 10⁻⁴
then based on the mole ratio, the moles of NH₃  = [(3.445 × 10⁻⁴) ÷ 3] × 2
                                                                                 = (1.14833 × 10⁻⁴) × 2
                                                      ∴ mole of NH₃   = 2.2967 × 10⁻⁴
                                                                                                                                       Now, # of molecules =  moles × Avogadro's Constant
         
∴ # of molecules of NH₃ = (2.2967 × 10⁻⁴)  ×  (6.02 × 10²³)
                                          =  1.383 × 10²⁰ molecules
       
      Thus, the molecules of NH₃ produced is ≈ 1.38 × 10²⁰ molecules.  
                                   

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<h2>Answer:</h2>

He is right that the energy of vaporization of 47 g of water s 106222 j.

<h3>Explanation:</h3>

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In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.

It means for vaporizing 18 g, 40.65 kJ energy is needed.

So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ

Hence the student is right about the energy of vaporization of 47 g of water.

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prisoha [69]

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Explanation:

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The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

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When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

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The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

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MA_775_DIABLO [31]

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Explanation:

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2 years ago
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