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Jet001 [13]
3 years ago
8

The equation for the complete combustion of ethane (C2H4) is

Chemistry
1 answer:
Mandarinka [93]3 years ago
5 0
-6x -8y = 28
6х + 4y =-28

would this be infinite many solutions or no solutions?
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A. Dalton's theory that atoms could not be divided was incorrect

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Which of the following elements has a complete outer shell of electrons? A. Iron (Fe) B. Hydrogen (H) C. Neon (Ne) D. Nitrogen (
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The answer is Neon

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Which option best characterizes the following scenario?
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Externalities are costs (negative externalities) or benefits (positive externalities), which are not reflected in free market prices. ... Market failure is a situation in which the free market leads to a misallocation of society's scarce ... and the killing of fish is not a cost that it would directly have to bear

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The pressure in a cylinder of nitrogen continuously decreases as a gas is released from it. On the other hand, a cylinder of pro
Juli2301 [7.4K]

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Explanation:

Nitrogen gas in nitrogen cylinder is in gaseous form only . So when gas is released , pressure decreases because moles of  gas in the cylinder decreases .

Propane gas in cylinder is in liquified form . The gas is partially in gaseous and liquid form . So when gas is released outside , pressure decreases inside but due to low pressure , some liquid form is converted into gaseous form which restores the pressure of gas inside cylinder . Hence , the pressure of gas remains unchanged inside cylinder.  

4 0
3 years ago
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

                               = (0.12 + 0.0116) mol

                                = 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

            = 3.37 (approx)

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
3 years ago
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