M.P of oleic acid = 13 deg C
<span>B.Pof oleic acid = 360 deg C </span>
<span>(360 - 13) / 100 = 3.47 </span>
<span>So 1 deg O = 3.47 deg C </span>
<span>The scale do not start at 0 deg C, it starts at 13 deg C. So to convert deg C to deg O,
subtract 13 then divide by 3.47 </span>
<span>deg O = (deg C - 13) / 3.47 </span>
<span>convert O to C multiply by 3.47 then add 13 </span>
<span>deg C = (deg O x 3.47) + 13 </span>
<span>convert 0 deg C to deg O </span>
<span>deg O = (0 deg C - 13) / 3.47 </span>
<span>= - 3.75 deg O</span>
Answer:
Explanation:
All you need to know is the atomic mass of platinum, and Avogadro's number.
2.00g Pt divided by atomic mass gives you the moles of platinum, and multiplying by avogadro's number (6.022 x 10^23) gives you the number of atoms.
Atomic mass of platinum can be found on any periodic table.
Hope this helped.
Answer:
F - O - S - Mg - Ba
Explanation:
as you move left to right on the periodic table the number of electrons increase.
Answer:
a. 1,500 kJ
b. 
Explanation:
Hello,
In the attached picture you will find the solution for the exercise.
- Take into account that ΔT=0.
Best regards.
Explanation:
The given data is as follows.
= 9,
= ?
Z for hydrogen = 1
As we know that,
Energy (E) = 
where, h = planck's constant =
Js
c = speed of light =
m/s
= wavelength
According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.
![\Delta E = -2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20-2.179%20%5Ctimes%2010%5E%7B-18%7D%20J%20%5Ctimes%20%28Z%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B1%7D%7D%5D)
or,
= ![-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]](https://tex.z-dn.net/?f=-2.179%20%5Ctimes%2010%5E%7B-18%7D%20J%20%5Ctimes%20%28Z%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B1%7D%7D%5D)
Putting the given values into the above equation as follows.
= ![-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]](https://tex.z-dn.net/?f=-2.179%20%5Ctimes%2010%5E%7B-18%7D%20J%20%5Ctimes%20%28Z%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B2%7D%7D%20-%20%5Cfrac%7B1%7D%7Bn%5E%7B2%7D_%7B1%7D%7D%5D)
=
n = 2
Thus, we can conclude that the final level of the electron is 2.