Answer:
<h2>Dog's mitochondria lack the transport protein that transport pyruvate ( end product of glycolysis) across the outer mitochondrial membrane
.</h2>
Explanation:
1. As given here that dog's mitochondria can use only fatty acids and also amino acids for their respiration, and as compared to others, Dong's cell produce more lactate then normal, this indicate that his mitochondrial membrane is different then others.
2. The aerobic phases of cellular respiration in eukaryotes occur within mitochondria. These aerobic phases are the TCA Cycle and the electron transport chain. Glycolysis occurs in the cytoplasm and the products of glycolysis enter into the mitochondria to continue cellular respiration.
3. These condition shows that dog's mitochondria lack the transport protein of mitochondria that moves pyruvate across the outer mitochondrial membrane.
Insect's or Bugs preserved in Amber
Answer:
0.712 mol
Explanation:
The easiest way to do this is to use a proportion.
1 mol of copper = 63.5 grams (check this using your periodic table).
x mol of copper = 45.2 grams
1/x = 63.5 / 45.2 Cross multiply
63.5 x = 1 * 45.2 Divide by 63.5
x = 45.2/63.5
x = 0.712 mol Answer to 3 sig digs
Answer:
Respiration
Explanation:
The act of breathing, simple as that!
Hope this helped!!
The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
