Answer:
The order of some common metals in the electromotive series, starting with the most easily oxidized, is: lithium, potassium, calcium, sodium, magnesium, aluminum, zinc, chromium, iron, cobalt, nickel, lead, hydrogen, copper, mercury, silver, platinum, and gold.
Explanation:
When palmitate (C16) completely oxidized, net yield is 129 ATP.
There are 7 β-oxidation cycles for palmitate to completely oxidazed.
One β-oxidation (beta-oxidation) cycle produces 1 NADH, 1 FADH2, and 1 acetyl CoA (sea the picture below).
For 7 cycles: 7 NADH, 7 FADH2, and 8 acetyl CoA.
NADH and FADH2 enter Electron Transport System (ETS) cycle to generate 3ATP and 2ATP respectively.
Acetyl CoA enter into tricarboxylic acid cycle (TCA) and each CoA gives 12 ATP.
7 NADH = 7 x 3 = 21 ATP
7 FADH2 = 7 x 2 = 14 ATP
8 acetyl-CoA = 8 x 12 = 96 ATP
Net yield of ATP = 21 ATP + 14 ATP + 96 ATP - 2 ATP (used during conversion of palmitic acid into palmitoyl CoA)
Net yield of ATP = 129 ATP
More about beta-oxidation: brainly.com/question/14993930
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Remember this:
atomic mass= atomic number + number of neutrons
The atomic number (same thing as number of protons) in this case is 15. The atomic mass is 31.
So, we subtract 15 from 31 to find the number of neutrons.
31-15=16
16 neutrons!!!
Answer:
7
Explanation:
Assume we have 1 L of each solution.
Solution 1
![\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}](https://tex.z-dn.net/?f=%5Ctext%7B%5BH%24%5E%7B%2B%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20H%7D%5E%7B%2B%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D)
Solution 2
pH = 12
pOH = 14.00 - pOH = 14.00 - 12 = 2.0
![\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%24%5E%7B-%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pOH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20OH%7D%5E%7B-%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D)
3. pH after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 10⁻² 10⁻²
C/mol: -10⁻² -10⁻²
E/mol: 0 0
The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.
pH = 7
In science you want to use observations and measurements.