The question is incomplete: the complete question is:
Rank molecules or group of molecules by when mammals use as fuel, starting right after meal to only through?
starvation.
a) glucose from glucose-2-phosphate from glycogen.
b) free glucose and amino acids
c) fatty acids from triglycerides
d) glucose produced from amino acids in liver.
Answer:
B
A
C
D
Explanation:
Free glucose and amino acid enters into the blood stream directly and can be used immediately after meals by mammals.
The liver stores excess glucose as glycogen, hence when glucose levels begin to reduce in the body, the pancreas releases glucagon which converts glycogen to glucose.
Triglycerides are broken down to give fatty acids which gives much more energy than glucose.
During starvation, amino acids in the liver is converted to glucose when a person goes for days without eating.
Answer:
17 kJ
Explanation:
Calculation for the Calculate the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C.
Using this formula
q = mC∆T
Where,
q represent Energy
m represent Mass of substance=0.60kg=600g
C represent Specific heat capacity=2.44J·g−1K−1.
∆T represent change in Temperature=2.2°C to 13.7°C.
Let plug in the formula
q=(0.60 kg x 1000 g/kg)(2.44 J/gº)(13.7°C-2.2°C)
q = (600g)(2.44 J/gº)(11.5º)
q=16.836 kJ
q= 17 kJ (Approximately)
Therefore the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C will be 17 kJ
The answer is 308 K.
The formula is C + 273.15 = K
this meaning the formula for this problem would be 35 + 273.15 = 308.15.
Out of all the option choices that would be rounded to 308K.
Answer:
2C3H8O + 9O2 ==> 6CO2 + 8H2O ... balanced equation
moles propanol = 5.26 g x 1 mol/60.1 g = 0.0875 moles
moles O2 = 31.8 g x 1 mol/31.9 g = 0.997 moles O2
Propanol is limiting based on the mol ratio in balance equation of 2 : 9
To find mass of O2 (excess reagent) left over, we will first find moles O2 used up.
moles O2 used = 0.0875 mol propanol x 9 mol O2/2 mol propanol = 0.394 moles O2 used
moles O2 left over = 0.997 mol - 0.394 mol = 0.603 mol O2 left
mass O2 left = 0.603 mol O2 x 32 g/mol = 19.3 g O2 left over
