<span>When the electron in a hydrogen atom transitions from a high energy state to a lower energy state, the energy lost from the electron is used to produce a photon corresponding to the loss of energy. That photon will correspond to exactly 1 wavelength. And since a hydrogen atom has only 1 electron, at any given moment, it can only produce 1 photon. And in order to simultaneously produce 4 photons for 4 spectral lines, that would require a simultaneous transition of 4 electrons which is 3 too many for a hydrogen atom.</span>
PKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5
![ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}](https://tex.z-dn.net/?f=ka%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BCH3CH2COO%5E-%5D%7D%7B%5BCH3CH2COOH%5D%7D%20)
![[CH3CH2COO^-] ](https://tex.z-dn.net/?f=%5BCH3CH2COO%5E-%5D%0A)
= 0.05
![[CH3CH2COOH]](https://tex.z-dn.net/?f=%5BCH3CH2COOH%5D)
= 0.10
Therefore 1.259x10^-5 =
![\frac{[H^+][0.05]}{[0.1]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5B0.05%5D%7D%7B%5B0.1%5D%7D%20)
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore
![[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%20%20%5Cfrac%7B%281.259%2A10%5E-5%29%280.1%29%7D%7B0.05%7D%20%20)
Therefore
![[H^+]= 2.513*10^-5](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%202.513%2A10%5E-5)
pH= -log [
] = -log(2.513*10^-5)= 4.59.
Answer:
15.3 %
Explanation:
Step 1: Given data
- Mass of the sample (ms): 230 g
- Mass of carbon (mC); 136.6 g
- Mass of hydrogen (mH): 26.4 g
- Mass of nitrogen (mN): 31.8 g
Step 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
Answer:
Cupid nitrate is what I'm going for
