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vitfil [10]
2 years ago
12

A sample of 0.900 mol N2O is placed in a sealed

Chemistry
1 answer:
Anettt [7]2 years ago
8 0

Answer:

  • <u>t = 284 min (rounded to 3 significant figures).</u>

Explanation:

<u>1. Decomposition reaction:</u>

Chemical equation:

  • 2N₂O → 2N₂ + O₂

<u>2. First order reaction:</u>

Rate law:

r=-\frac{d[N_2O]}{dt}= k[N_2O]\\ \\ \frac{d[N_2O]}{dt}= -k[N_2O]

Integrated rate law:

\frac{d[N_2O]}{[N_2O]}=-kdt\\ \\ ln[N_2O]-ln[N_2O]_0=-kt

<u>3. Determine the rate constant, k</u>

  • t = 42.0 min, [N₂O] = 0.640 mol

  • [N₂O]₀ = 0.900

  • ln(0.640)-ln(0.900)=-k(42.0min)

  • k = 0.00812 min⁻¹

<u>4. Time taken for the reaction to be 90.0% complete</u>

  • [N₂O] = 0.100[N₂O]=0.100×0.900 mol = 0.0900 mol

  • ln(0.0900) - ln(0.900) = - 0.00812 min⁻¹ × t

  • t = (2.30 / 0.00812) min ≈ 283.6 min ≈ 284 min (rounded to 3 significant figures).
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Answer:

Explanation:

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2 years ago
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2 years ago
At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?
algol [13]

Answer:

  • 820 K

Explanation:

As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.

  • KE α T.

Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

  • First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

  • Second, convert 0°C to kelvin:

         T (K) = T(°C) + 273.15 K= 273.15 K

Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

7 0
2 years ago
Water at 4.4°C is to flow through a horizontal, commercial steel pipe having a length of 305 m at the rate of 150 gal/min. If a
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Answer:

d = 70.5 mm

Explanation:

given,

length of pipe = 305 m

discharge rate = 150 gal/min

pipe diameter = ?

1 gal/min = 6.30902 ×  10⁻⁵ m³/s

150 gal/min = 150 × 6.30902 ×  10⁻⁵ m³/s

                   = 9.46 × 10⁻³ m³/s

h = \dfrac{flv^2}{2gd}

h = \dfrac{flv^2}{2gd}

Q = A V

h = \dfrac{fl(\dfrac{Q}{A})^2}{2gd}

h = \dfrac{fl(\dfrac{Q}{\dfrac{\pi}{4}d^2})^2}{2gd}

h= \dfrac{8flQ^2}{\pi^2gd^5}

f = 0.048 from moody chart using P/D = 0.00015

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7 0
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Answer:

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3) Some oxidation reaction.

Endothermic reaction: In endothermic reaction, energy is taken in from the surrounding, and the surrounding temperature decreases, this is known as endothermic reaction. In other words energy enters in endothermic reaction. Some example of exothermic reactions are:

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