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Blizzard [7]
3 years ago
14

How many moles of tetracycline (C₂₂H₂₄N₂O₈) are in 71.9 grams

Chemistry
1 answer:
slavikrds [6]3 years ago
4 0

Explanation:

<em>to </em><em>find </em><em>the </em><em>number</em><em> </em><em>of </em><em>mol</em><em>e</em><em>s</em><em> </em><em>for </em><em>tetracycline</em><em> </em><em>you </em><em>use </em><em>the </em><em>formula,</em>

<em>number </em><em>of </em><em>moles=</em><em>mass/</em><em>molar </em><em>mass</em>

<em>the </em><em>mass </em><em>is </em><em>given</em><em> </em><em>in </em><em>the </em><em>question</em><em> </em><em>which </em><em>is </em><em>7</em><em>1</em><em>.</em><em>9</em><em>g</em><em> </em><em>and </em><em>the </em><em>molar </em><em>mass </em><em>is </em><em>4</em><em>4</em><em>4</em><em>.</em><em>4</em><em>3</em><em>(</em><em>you </em><em>find </em><em>I </em><em>by </em><em>adding</em><em> </em><em>the </em><em>number </em><em>of </em><em>atoms</em><em> </em><em>for </em><em>each </em><em>element </em><em>in </em><em>the </em><em>compound</em><em>,</em><em>1</em><em>2</em><em>×</em><em>2</em><em>2</em><em>+</em><em>1</em><em>×</em><em>2</em><em>4</em><em>+</em><em>1</em><em>4</em><em>×</em><em>2</em><em>+</em><em>1</em><em>6</em><em>×</em><em>8</em><em>)</em>

<em>n=</em><em>7</em><em>1</em><em>.</em><em>9</em><em>g</em><em>/</em><em>4</em><em>4</em><em>4</em><em>.</em><em>4</em><em>3</em>

<em>=</em><em>0</em><em>.</em><em>1</em><em>6</em><em>g</em><em>/</em><em>mol</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

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                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

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Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
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When E° cell is an electrochemical cell which comprises of two half cells.
 
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when we have the balanced equation of this half cell :

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and we have  also this balanced equation of this half cell :

Ag+(aq)  + e- → Ag(s)  and E°2 = 0.8 V 

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

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∴ the correct answer is 2.46 V




6 0
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