Answer:
uh.. How many do you want done??
Explanation:
<u> electrical energy to chemical energy</u>
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.
Answer:5-3=2
As 5apple eaten 3apple so left 2apple
B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.
Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P).