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Inessa05 [86]
3 years ago
6

What is the molecularity of this elementary step?

Chemistry
2 answers:
xxTIMURxx [149]3 years ago
7 0

Answer:

<u><em>The answer is</em></u>: <u>B) bimolecular.</u>

<u />

Explanation:

Molecularity is the number of molecules that are part as reagents in an elementary process,<em> that is, the sum of the molecules of each reagent before forming the activated complex to become the products. </em>

<u><em>Bimolecular process</em></u>: <u>A + B ---> C </u>

<u><em>The answer is</em></u>: <u>B) bimolecular.</u>

dybincka [34]3 years ago
3 0
The correct answer is b. bimolecular
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topjm [15]

Answer:

8.37g

Explanation:

Step 1 :

The balanced equation for the reaction. This is given below:

N2 + 2O2 —> 2NO2

Step 2:

Data obtained from the question.

Volume (V) of N2 = 2L

Pressure (P) = 840mmHg

Temperature (T) = 24°C

Number of mole (n) of N2 =?

Step 3:

Conversion to appropriate unit.

For pressure:

760mmHg = 1atm

840mmHg = 840/760 = 1.11 atm

For Temperature:

T(K) = T(°C) + 273

T(°C) = 24°C

T(K) = 24°C + 273

T(K) = 297K

Step 4:

Determination of the number of mole N2.

The number of mole of N2 can be obtained by using the ideal gas equation as follow:

Volume (V) of N2 = 2L

Pressure (P) = 1.11 atm

Temperature (T) = 297K

Number of mole (n) of N2 =?

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

Divide both side by RT

n = PV / RT

n = 1.11 x 2 / 0.082 x 297

n = 0.091 mole

Therefore, the number of mole of N2 that reacted is 0.091 mole

Step 5:

Determination of the mass of NO2 produced from the reaction. This is illustrated below:

N2 + 2O2 —> 2NO2

From the balanced equation above,

1 mole of N2 produced 2 moles of NO2.

Therefore, 0.091 mole of N2 will produce = 0.091 x 2 = 0.182 mole of NO2.

Finally, we will convert 0.182 mole of NO2 to gram as shown below:

Number of mole NO2 = 0.182 mole

Molar mass of NO2 = 14 + (16x2) = 46g/mol

Mass = number of mole x molar mass

Mass of NO2 = 0.182 x 46

Mass of NO2 = 8.37g

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Both protons and neutrons (and their anti-particles) froze out:
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6 0
3 years ago
The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

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3 years ago
When the mixture was moved from room temperature to a higher temperature, the mixture turned darker brown. Which of the followin
Wittaler [7]

Answer:

The equilibrium between the two forms of the gas is disturbed at high temperatures.

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