Answer:
c. By itself, heme is not a good oxygen carrier. It must be part of a larger protein to prevent oxidation of the iron.
e. Both hemoglobin and myoglobin contain a prosthetic group called heme, which contains a central iron ( Fe ) (Fe) atom.
f. Hemoglobin is a heterotetramer, whereas myoglobin is a monomer. The heme prosthetic group is entirely buried within myoglobin.
Explanation:
The differences between hemoglobin and myoglobin are most important at the level of quaternary structure. Hemoglobin is a tetramer composed of two each of two types of closely related subunits, alpha and beta. Myoglobin is a monomer (so it doesn't have a quaternary structure at all). Myoglobin binds oxygen more tightly than does hemoglobin. This difference in binding energy reflects the movement of oxygen from the bloodstream to the cells, from hemoglobin to myoglobin.
Myoglobin binds oxygen
The binding of O 2 to myoglobin is a simple equilibrium reaction:
Answer:
This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol
Explanation:
The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.
Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L
Answer:
12.32 L.
Explanation:
The following data were obtained from the question:
Mass of CH4 = 8.80 g
Volume of CH4 =?
Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:
Mass of CH4 = 8.80 g
Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol
Mole of CH4 =?
Mole = mass/Molar mass
Mole of CH4 = 8.80 / 16
Mole of CH4 = 0.55 mole.
Finally, we shall determine the volume of the gas at stp as illustrated below:
1 mole of a gas occupies 22.4 L at stp.
Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.
Thus, 8.80 g of CH4 occupies 12.32 L at STP.
