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Butoxors [25]
3 years ago
9

What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)

8.68 3) 5.32 4) 2.31 5) < 2.0
Chemistry
1 answer:
blsea [12.9K]3 years ago
6 0
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

                    =0.00516 M 

Thus, pOH = 2.30 
 pH = 14 - pOH = 11.7
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Extra significant figures are kept to avoid round-off errors.

We then calculate the moles of the organic compound:

(0.6654 g)(mol/46.07) = 0.0144432 mol

We then calculate the heat released per mole and convert to the proper units. (The conversion between kJ and J is infinitely precise and is not involved in the consideration of significant figures)

(19986.264 J)(1kJ/1000J) / (0.0144432 mol) = 1384 kJ/mol

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Why is dissolving salt in water is a physical change? Help me please
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Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/
Ivenika [448]

Answer:

The answer to your question is:   C₃H₃O  This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

                                     44 g -------------------- 12 g

                                     3.2007 g ------------    x

                                      x = (3.2007 x 12) / 44

                                      x = 0.8729 g of Carbon

                                     12 g of C --------------  1 mol

                                     0.8729 g --------------  x

                                     x = (0.8729 x 1) / 12

                                     x = 0.0727 mol of Carbon

For Hydrogen

                                 18 g ---------------------- 1 g

                              1.3102 g -------------------  x

                                   x = (1.3102 x 1) / 18

                                  x = 0.0727 g of Hydrogen                                      

                                  1 g ------------------------ 1 mol

                                  0.0727g ----------------  x

                                  x = (0.0727 x 1)/1

                                  x = 0.0727 mol of Hydrogen

For oxygen

                    g of Oxygen = g of sample - g of Carbon - g of hydrogen

                    g of Oxygen = 1.3109 - 0.8709 - 0.0727

                    g of Oxygen = 0.3673

                                  16 g of Oxygen ------------- 1 mol of O

                                  0.3673 g ---------------------   x

                                   x = (0.3673 x 1)/ 16

                                   x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon              0.0727 / 0.023  = 3.1  ≈ 3

Hydrogen         0.0727 / 0.023 = 3.1  ≈ 3

Oxygen             0.0230 / 0.023 = 1

                                        C₃H₃O

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