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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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How do you find the amount of moles is .032 grams of water and whats the answer
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Answer:

\boxed {\boxed {\sf 0.0018 \ mol \ H_2 O }}

Explanation:

First, we need to find the molecular mass of water (H₂O).

H₂O has:

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  • 1 Oxygen atom (implied subscript of 1)

Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.

  • Hydrogen: 1.0079 g/mol
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There are 2 hydrogen atoms, so multiply the mass by 2.

  • 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol

Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.

  • 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol

Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.

0.032 \ g  \ H_2 O* \frac{ 1 \ mol \ H_2 O}{18.0152 \ g \ H_2 O}

Multiply. The grams of H₂O will cancel out.

0.032 * \frac{1 \ mol \ H_2 O}{18.0152 }

\frac{0.032 *1 \ mol \ H_2 O}{18.0152 }

0.00177627781 \ mol \ H_2 O

The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.

The 7 in the hundred thousandth place tells us to round up.

0.0018 \ mol \ H_2 O

There are about <u>0.0018 moles in 0.032 grams.</u>

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Radioactive radium has a half-life of approximately 1,599 years. The initial quantity is 13 grams. How much (in grams) remains a
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The quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

<h3>What is half life period? </h3>

The time taken by substance to reduce to its half of its initial concentration is called half life period.

We will use the half- life equation N(t)

N e^{(-0.693t) /t½}

Where,

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By substituting all the value, we get

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learn more about half life period:

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