Answer:
Whales facilitate carbon absorption in two ways. On the one hand, their movements — especially when diving — tend to push nutrients from the bottom of the ocean to the surface, where they feed the phytoplankton and other marine flora that suck in carbon, as well as fish and other smaller animals.
The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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Supercooling<span>, a state where liquids do not solidify even below their normal freezing point. Means sometimes we have liquid water below 0 degree C.</span>
Answer:
A
Explanation:
Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

Therefore, from the chemical equation, we have that:
![\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%28-317%5Ctext%7B%20kJ%2Fmol%7D%29%20%3D%20%5Cleft%5B%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%2B%20%20%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20H%24_2%24O%7D%20%20%5Cright%5D%20%20%20-%5Cleft%5B3%20%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20H%24_2%24%7D%2B%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20N%24_2%24O%7D%5Cright%5D%20%5Cend%7Baligned%7D)
Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:
![\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%28-317%5Ctext%7B%20kJ%2Fmol%7D%29%20%26%20%3D%20%5Cleft%5B%20%5CDelta%20H%5E%5Ccirc%20_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%2B%20%28-285.8%5Ctext%7B%20kJ%2Fmol%7D%29%5Cright%5D%20-%5Cleft%5B%203%280%29%20%2B%20%2882.1%5Ctext%7B%20kJ%2Fmol%7D%29%5Cright%5D%20%5C%5C%20%5C%5C%20%5CDelta%20H%5E%5Ccirc%20_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%26%20%3D%20%28-317%20%2B%20285.8%20%2B%2082.1%29%5Ctext%7B%20kJ%2Fmol%7D%20%5C%5C%20%5C%5C%20%26%20%3D%2050.9%5Ctext%7B%20kJ%2Fmol%7D%20%5Cend%7Baligned%7D)
In conclusion, our answer is A.