Answer:
Explanation:
Molecular FormulaC4H7O2
Average mass87.098 Da
Monoisotopic mass87.045151 Da
Answer:
146.3g NaCl (mol NaCl/58.44g NaCl) = 2.50 mol NaCl
1.5M NaCl = 1.5 mol NaCl / 1 L = 2.5 mol NaCl / x L, solve for x
x L = 2.5 mol NaCl / 1.5 mol NaCl = 1.66 L
It gives the answer and all the working.
To put it another way:
Dividing the amount required by the molar mass
we quickly see that 2.5 moles are required.
One litre of 1.5 molar solution gives 1.5 moles
we need a further mole, which is 2/3 of 1.5 so 2/3 of a litre.
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.
Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
Answer: b
explanation: i had the question on a test and got it right.
