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Monica [59]
3 years ago
11

The first-order decomposition of n2o at 1000 k has a rate constant of 0.76 s-1. if the initial concentration of n2o is 10.9 m, w

hat is the concentration of n2o after 9.6 s
Chemistry
1 answer:
Reptile [31]3 years ago
3 0
For a first-order reaction, the rate law would be expressed as:<span>

r = dC / dt = -kC

Integrating it from time zero and the initial concentration, Co, to time, t, and the final concentration, C. We will obtain the first-order integrated law as follows:

ln C/Co= -kt

To determine the concentration of N2O in the system at a certain time, we simply substitute the given values from the problem statement as follows:

</span>ln C / Co = -kt<span>
ln C / 10.9 = -0.76 (9.6)
e^ln C / 10.9 = e^-0.76 (9.6)
C / 10.9 = 6.78 x 10^-4
C= 7.39 x 10^-3 m
<span>
Therefore, the concentration of N2O in the system after 9.6 s would be 7.39 x10^-3 m.
</span></span><span><span>
</span></span>
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Draw the major resonance contributors for the enamine shown. do not draw resonance and curved arrows. include formal charges and
Makovka662 [10]

<em>Answer:</em>

There are two major resonance contributors for the en-amine. One contributor have no formal charges, only have one lone pair at N while the other one has positive and negative formal charges.

<em>Explanation:</em>

An en-amine is formed by the condensation reactions of aldehydes or ketone with secondary amine.

The contributor<em> </em>I have no formal charges. It has only one lone pair at Nitrogen atom.

The contributor II has +1 formal charges at N, and -1 formal charges at α Carbon.

Please see attachment. The resonance contributor are given.


Download docx
5 0
3 years ago
A stock solution is 28.2 percent ammonia (NH3) by mass, and the solution has a density of 0.8990 grams per milliliter. What volu
Savatey [412]

Answer:

\boxed{\text{20.2 mL}}

Explanation:

Assume that the volume of the stock solution is 1 L.

1. Mass of  stock solution

\text{Mass} = \text{1000 mL} \times \dfrac{\text{0.8990 g}}{\text{1 mL}} = \text{899.0 g}

2.Mass of NH₃

\text{Mass of NH}_{3} = \text{899.0 g stock} \times \dfrac{\text{28.2 g NH}_{3}}{\text{100 g stock}} = \text{253.5 g NH}_{3}

3. Moles of NH₃

\text{Moles of NH}_{3} = \text{253.5 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.89 mol NH}_{3}

4. Molar concentration of stock solution

c = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{14.89 mol}}{\text{1 L}} = \text{14.89 mol/L}

5. Volume of stock needed for dilution

Now that you know the concentration of the stock solution, you can use the dilution formula .

c_{1}V_{1} = c_{2}V_{2}

to calculate the volume of stock solution.

Data:

c₁ = 14.89   mol·L⁻¹; V₁ = ?

c₂ = 0.500 mol·L⁻¹; V₂ = 600 mL

Calculations:

\begin{array}{rcl}14.89V_{1} & = & 0.500 \times 600\\14.89V_{1} & = & 300\\V_{1} & = & \text{20.2 mL}\\\end{array}\\\text{You will need $\boxed{\textbf{20.2 mL}}$ of the stock solution.}

4 0
3 years ago
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K
Ilya [14]

Answer:

-0.050 kJ/mol.K

Explanation:

  • A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
  • The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K

All in all, ΔG° = 0 at 400. K.

We can find ΔS° using the following expression.

ΔG° = ΔH° - T.ΔS°

0 = -20 kJ/mol - 400. K .ΔS°

ΔS° = -0.050 kJ/mol.K

8 0
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Which best compares DNA and RNA with regard to the process of protein production? DNA transforms from a nucleic acid into a prot
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Answer:

its d

Explanation:i got it right on egde

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3 years ago
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Kinetic energy is a measure of a particles what?
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B is the answer because it is
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