Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
1-energy
2- force
3- force
4- force
5- energy
6- energy
The following are the statements, which indicates that the fossil record suggests about evolution on Earth:
1. Humans have only recently existed on Earth.
2. The organisms originally lived only in Earth's water.
3. The plants did not have flowers when dinosaurs existed.
The fossil remains have been discovered in the rocks of all the ages. The simplest organisms fossils are witnessed in the oldest rocks, and the fossils of more composite species are found in the newest rocks. This supports the theory of evolution suggested by Darwin, according to which the simple life forms slowly gets evolved into more composite ones.
Fe needs to have a positive charge of +3 to balance out -3 Cl
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.
