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liberstina [14]
3 years ago
13

The decomposition of nitramide, O2NNH2, in water has the chemical equation and rate law O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H

+] A proposed mechanism for this reaction is O2NNH2(aq)⥫⥬=k−1k1O2NNH−(aq)+H+(aq)(fast equilibrium) O2NNH−(aq)−→k2N2O(g)+OH−(aq)(slow) H+(aq)+OH−(aq)−→k3H2O(l)(fast) What is the relationship between the observed value of k and the rate constants for the individual steps of the mechanism?
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law

Explanation:

Please, observe the solution in the attached Word document.

Download docx
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Be careful because answer A has same masses but different atomic numbers so different atoms(elements)!!!
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The element nickel has the following properties. Select the two that are chemical properties of nickel.
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Ionic compounds can conduct electricity when dissolved in water due to the presence of free floating ions.
tankabanditka [31]

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True

Explanation:

8 0
3 years ago
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The reactants of two chemical equations are listed. Equation 1: PbNO3 + Mg Equation 2: PbNO3 + KI Based on the type of reaction,
Vika [28.1K]

Answer:

Equation 2, because K being more reactive, exchanges position with Pb in PbNO3.

Explanation:

Hello there!

In this case, according to the given reactions, it is possible to realize that according to the reactivity series, since K is is group 1A of alkali metals, we infer it is by far more reactive than magnesium, for that reason last two choices can be easily discarded. Now, considering equation 2, it would be necessary to complete it to figure out the correct option:

Pb(NO_3)_2 + KI \rightarrow PbI_2+KNO_3

Whereas it can be seen that potassium exchanges position with Pb according to the double displacement reaction; therefore, the correct answer is "Equation 2, because K being more reactive, exchanges position with Pb in PbNO3".

Best regards!

8 0
3 years ago
The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
Valentin [98]

Answer:

(a)

0.0342M

(b)

t_{1/2}=17.36s\\t_{1/2}=23.15s

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\

[NOBr]=\frac{1}{29.2/M}=0.0342M

(b) Now, for a second-order reaction, the half-life is computed as shown below:

t_{1/2}=\frac{1}{k[NOBr]_0}

Therefore, for the given initial concentrations one obtains:

t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.072M}=17.36s\\t_{1/2}=\frac{1}{\frac{0.80}{M*s}*0.054M}=23.15s

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