In this question, you are given the average cofactor mass per cell (41.5pg) and the total cells count(105 cells). You are asked how much cofactor that will be found from those cells(microgram= 10^6 picogram). Then the calculation would be:
Cofactor mass= cofactor per cell * cell count= 41.5pg/cell * 105 cells= 4357.5pg= 4.36 x 10^3pg
Then convert the picogram(pg) into microgram: 4.36 x 10^3pg/ (10^6pg/microgram)= 4.36x10^-3 microgram or 0.00436 microgram
if 105 cells mean 10^5 cells, the answer should be 4.15 microgram
The First Ionization energy of Nitrogen is greater (Not smaller)than that of Phosphorous. This is because going down the group (N and P are in same group) the number of shells increases, the distance of valence electrons from Nucleus increases and hence due to less interaction between nucleus and valence electrons it becomes easy to knock out the electron.
<span>The second ionization energy of Na is larger than that of Mg because after first loss of electron Na has gained Noble Gas Configuration (Stable Configuration) and now requires greater energy to loose both second electron and Noble Gas Configuration. While Mg after second ionization attains Noble Gas Configuration hence it prices less energy.</span>
Hello!
To solve this problem we are going to use the
Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is
4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is
5,75Have a nice day!
So what we know:
-Atomic Mass = Protons + Neutrons
-Atomic Number is the number of protons
Magnesium's atomic number is 12, so the natural occurring isotope for magnesium is Mg-12 (12 protons and 12 neutrons). Added up we have an atomic mass of 24 amu. Which means if we added one neutron in Mg-13, our atomic mass would be 25 amu.
We can use the equation:
(amu of isotope 1)x + (amu of isotop 2)(x-1) = Average atomic mass
where isotope 1 is the fractional abundance we're solving for.
Plugged in it looks like this:
24x + 25(1-x) = 24.3
Now to solve for x:
24x + 25 - 25x = 24.3
-x + 25 = 24.3
-x = -.7
x = .7
So in this case, the fractional abundance of Mg-12 would be .7, or 70%.<span />
The transition metal with the smallest atomic mass is Scandium (Sc).
Hope this helps~
