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castortr0y [4]
2 years ago
14

How many km are in 5.6mm? 5.6x103 5.6x10-6 5.6x10-3 5.6x106​

Chemistry
1 answer:
il63 [147K]2 years ago
7 0

Hey there :)

<em>Q</em><em>u</em><em>e</em><em>s</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em> </em><em>How many km are in 5.6mm? </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>3 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-6 </em>

<em>=</em><em>></em><em>5.6x10</em><em>^</em><em>-3 </em>

<em>=</em><em>></em><em> </em><em>5.6x10</em><em>^</em><em>6</em>

<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em>-</em>

5.6 \times 10^{ - 6}

<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em><em>-</em>

By using the formula-

1millimeter =  \frac{1}{1000000}

As 1 with 6 zeros, we convert it into exponential form.

=  >  \frac{1}{10^{6} }

As this above value is fraction type, we can do the reciprocal, thus, the exponent gets a negative value.

=  > 10^{ - 6}

Now combine with given question.

=  > 5.6 \times 10^{ - 6}

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Explanation:

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2 years ago
Write the balanced chemical equation for the reaction shown here. answer chemistry
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I found the attached image with the same statement of your question and think it may be very useful for you that I use it to show how to answer this question (furthermore I think it may be the same reaction that you forgot to include).

As you can see, there are one image on the left side and other image on hte right side of the figure.

Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.

Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:

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Right side:

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I hope this is useful for you..






7 0
3 years ago
A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
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Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

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