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xz_007 [3.2K]
3 years ago
9

How many moles of n are in 0.163 g of n2o?

Chemistry
2 answers:
slava [35]3 years ago
7 0

Answer:

n_N=7.41x10^{-3}molN

Explanation:

Hello,

In this case, one develops the following mass-mole relationship, taking into account that N_2O has two nitrogens inside, thus:

n_N=0.163gN_2O*\frac{1molN_2O}{44gN_2O}*\frac{2molN}{1molN_2O}\\n_N=7.41x10^{-3}molN

Best regards.

zimovet [89]3 years ago
3 0

First convert 0.163 grams of N2O to mol by dividing it with the molecular weight. The molecular weight of N2O is 44 grams/mol. The answer would be 3,79x10^-3. Then multiply it with 2 since there are 2 Nitrogen in one mole of N2O. Therefore, there are 7.41x10^-3 moles of Nitrogen. 

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Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

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