Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.
Explanation :
To calculate the percentage composition of element in sample, we use the equation:
Given:
Mass of carbon = 1.94 g
Mass of hydrogen = 0.48 g
Mass of sulfur = 2.58 g
First we have to calculate the mass of sample.
Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur
Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g
Now we have to calculate the percentage composition of a compound.
Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.
Answer:
A
Explanation:
hail will fall push it back up making it bigger comes back down
E = mc^2
E = 8.90 * 10^12 Joules
c = 3 * 10^8 m/s
m = ????
8.90 * 10^12 = m * (3 * 10^8)^2
8.90 * 10^12 = m * 9 * 10^16
9.889 * 10^-4 kg = m <<<<< answer
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.
The last one will be It's better to use the average data instead of the single trail because the average you don't have to multiply add subtract and all that nonsense but if you use the single trail you will have to do all the adding multiplying etc.
