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Karolina [17]
3 years ago
6

Ethyne (C2 H2 (g), mc032-1.jpgHf = 226.77 kJ/mol) undergoes complete combustion in the presence of oxygen to produce carbon diox

ide (CO2 (g), mc032-2.jpgHf = –393.5 kJ/mol ) and water (H2 O(g), mc032-3.jpgHf = –241.82 kJ/mol) according to the equation below.
mc032-4.jpg

What is the enthalpy of combustion (per mole) of C2 H2 (g)?
–2511.2 kJ/mol
–1255.6 kJ/mol
–862.1 kJ/mol
–431.0 kJ/mol
Chemistry
2 answers:
Gnesinka [82]3 years ago
3 0
2 C2H2 + 5 02  >  4 CO2 + 2 H2O

Products  - Reactants ( all units are kJ/mo1):

(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1

-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1

answer: -1255.6 kJ/mo1
GuDViN [60]3 years ago
3 0

Answer:

B. –1255.6 kJ/mol

Explanation:

this is correct on ed-genuity, hope this helps! :)

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15. A substance that has a melting point of 1074 K
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Answer:

Ionic solid.

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Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat
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Explanation:

According to the given data, the turnover number can be calculated as follows.

      Turnover number = K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}

     V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}

Therefore, moles of CO_{2} hydrolyzed is as follows.

Moles of CO_{2} hydrolyzed = \frac{Mass of CO_{2}}{Molar mass of CO_{2}}

                 = \frac{0.30}{44}

                 = 0.00682 moles

Now, moles of CO_{2} hydolyzed per second is calculated as follows.

Moles of CO_{2} hydolyzed per second = \frac{0.00682}{60}

             = 1.137 \times 10^{-4} moles/second = V_{max}

And,

Moles of enzyme = \frac{Mass}{\text{Molar mass}}

                       = \frac{10.0 \mu g}{30000}

                       = 3.33 \times 10^{-10} moles

Therefore, the value of K_{cat} is as follows.

    K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}

               = 0.3411 \times 10^{6} per second

               = 0.3411 \times 60 \times 10^{6} per minute

               = 20.466 \times 10^{6} per minute

Thus, we can conclude that the turnover number (K_{cat}) of carbonic anhydrase (in units of min^{-1}) is 20.466 \times 10^{6} per minute.

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