The change concentration of the product of a reaction over time is best shown by line graph.
You use the vertical axis (normally y-axis) to show the concentration of the product and the horizontal axis (normally x-axis) to show the time.
In that way the several points may show a tendency that you can show as a line, and the slope of the line is the change of concentration divided by the time.
When 2.50 g is burned then in oxygen then 1.25kj of heat is produced.
Answer:
Explanation:
From the given information:
The energy of photons can be determined by using the formula:
![E = \dfrac{hc}{\lambda}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bhc%7D%7B%5Clambda%7D)
where;
planck's constant (h) = ![6.63 \times 10^ {-34}](https://tex.z-dn.net/?f=6.63%20%5Ctimes%2010%5E%20%7B-34%7D)
speed oflight (c) = ![3.0 \times 10^8 \ m/s](https://tex.z-dn.net/?f=3.0%20%5Ctimes%2010%5E8%20%5C%20m%2Fs)
wavelength λ = 58.4 nm
![E = \dfrac{6.63 \times 10^{-34} \ J.s \times 3.0 \times 10^8 \ m/s}{58.4 \times 10^{-9 } \ m}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B6.63%20%5Ctimes%2010%5E%7B-34%7D%20%5C%20J.s%20%5Ctimes%203.0%20%5Ctimes%2010%5E8%20%5C%20m%2Fs%7D%7B58.4%20%5Ctimes%2010%5E%7B-9%20%7D%20%5C%20m%7D)
![E =0.34 \times 10^{-17} \ J](https://tex.z-dn.net/?f=E%20%3D0.34%20%5Ctimes%2010%5E%7B-17%7D%20%5C%20J)
![E = 3.40 \times 10^{-18 } \ J](https://tex.z-dn.net/?f=E%20%3D%203.40%20%5Ctimes%2010%5E%7B-18%20%7D%20%5C%20J)
To convert the energy of photon to (eV), we have:
![1 eV = 1.602 \times 10^{-19} \ J](https://tex.z-dn.net/?f=1%20eV%20%3D%201.602%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20J)
Hence
![3.40 \times 10^{-18 } \ J = \dfrac{1 eV}{1.602 \times 10^{-19 } \ J }\times 3.40 \times 10^{-18 } \ J](https://tex.z-dn.net/?f=3.40%20%5Ctimes%2010%5E%7B-18%20%7D%20%5C%20J%20%3D%20%5Cdfrac%7B1%20eV%7D%7B1.602%20%5Ctimes%2010%5E%7B-19%20%7D%20%5C%20%20J%20%20%7D%5Ctimes%203.40%20%5Ctimes%2010%5E%7B-18%20%7D%20%5C%20J)
![E = 2.12 \times 10 \ eV](https://tex.z-dn.net/?f=E%20%3D%202.12%20%5Ctimes%2010%20%5C%20eV)
E = 21.2 eV
b)
The equation that illustrates the process relating to the first ionization is:
![Hg_{(g)} \to Hg^+ _{(g)} + e^-](https://tex.z-dn.net/?f=Hg_%7B%28g%29%7D%20%5Cto%20Hg%5E%2B%20_%7B%28g%29%7D%20%2B%20e%5E-)
c)
The 1st ionization energy (I.E) of Hg can be calculated as follows:
Recall that:
I.E = Initial energy - Kinetic Energy
I₁ (eV) = 21.2 eV - 10.75 eV
I₁ (eV) = 10.45 eV
Since ;
![1 eV = 1.602 \times 10^{-19} \ J](https://tex.z-dn.net/?f=1%20eV%20%3D%201.602%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20J)
∴
![10.45 \ eV = \dfrac{1.602 \times 10^{-19 } \ J }{ 1 \ eV}\times 10.45 \ eV](https://tex.z-dn.net/?f=10.45%20%5C%20eV%20%3D%20%5Cdfrac%7B1.602%20%5Ctimes%2010%5E%7B-19%20%7D%20%5C%20J%20%20%7D%7B%201%20%5C%20eV%7D%5Ctimes%2010.45%20%5C%20eV)
Hence; the 1st ionization energy of Hg atom = ![1.67 \times 10^{-18} \ J](https://tex.z-dn.net/?f=1.67%20%5Ctimes%2010%5E%7B-18%7D%20%5C%20J)
![1.67 \times 10^{-21} \ kJ](https://tex.z-dn.net/?f=1.67%20%5Ctimes%2010%5E%7B-21%7D%20%5C%20kJ)
Finally;
![I_1 \ of \ the \ Hg (kJ/mol) = \dfrac{1.67 \times 10^{-21 \ kJ} \times 6.02 \times 10^{23} \ Hg \ atom }{1 \ Kg \ atom }](https://tex.z-dn.net/?f=I_1%20%5C%20of%20%5C%20the%20%5C%20Hg%20%28kJ%2Fmol%29%20%3D%20%5Cdfrac%7B1.67%20%5Ctimes%2010%5E%7B-21%20%5C%20kJ%7D%20%5Ctimes%206.02%20%5Ctimes%2010%5E%7B23%7D%20%5C%20Hg%20%5C%20atom%20%7D%7B1%20%5C%20Kg%20%5C%20atom%20%7D)
![\mathbf{= 1.005 \times 10^3 \ kJ/mol}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%201.005%20%5Ctimes%2010%5E3%20%5C%20kJ%2Fmol%7D)
Answer:
The OH group
Explanation:
Benzhydrol contains OH hydroxyl group in its molecule while fluorene does not. At first glance, one would think that OH, which contributes to hydrogen bonding would causes melting point of benzhydrol to be higher than fluorene. <em>However, </em>the structure of benzhydrol, which is 2 benzene rings connected to center hydroxyl carbon (PhCOHPh), allows for each benzene rings in benzhydrol to rotate until both rings are perpendicular to minimize repulsive force. This prevents the molecule from stacking on each other due to its non flat shape, and thus, lowering its melting point in contrast to flat fluorene molecule.