Answer:
0.7μM = 0.6 μM = 0.5 μM > 0.4 μM > 0.3 μM > 0.2 μM
Explanation:
An enzyme solution is saturated when all the active sites of the enzyme molecule are full. When an enzyme solution is saturated, the reaction is occurring at the maximum rate.
From the given information, an enzyme concentration of 1.0 μM Y can convert a maximum of 0.5 μM AB to the products A and B per second means that a 1.0 M Y solution is saturated when an AB concentration of 0.5 M or greater is present.
The addition of more substrate to a solution that contains the enzyme required for its catalysis will generally increase the rate of the reaction. However, if the enzyme is saturated with substrate, the addition of more substrate will have no effect on the rate of reaction.
<em>Therefore the reaction rates at substrate concentrations of 0.7μM, 0.6 μM, and 0.5 μM are equal. But the reaction rate at substrate concentrations of 0.2 μM is lower than at 0.3 μM, 0.3 μM is lower than 0.4 μM and 0.4 μM is lower than 0.5 μM, 0.6 μM and 0.7 μM.</em>
Answer:
4.70 moles of water have mass of 84.6g
Explanation:
Given data:
Number of moles of water = 4.70 mol
Mass in gram = ?
Solution:
Formula:
Mass = number of moles × molar mass
Molar mass of water is 18 g/mol
Mass = 4.70 mol × 18 g/mol
Mass = 84.6 g
Thus, 4.70 moles of water have mass of 84.6g.
Answer:
The answer is 0.75M HCl
Explanation:
To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:
To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:
C1V1 = C2V2
C1 = concentration of acid
C2 = concentration of base
V1 = volume of acid
V2 = volume of base
From the information supplied in the question:
concentration of acid (HCl) is the unknown
volume of acid (HCl) = 10ml
concentration of base (NaOH) = 0.15M
volume of base (NaOH) = 50ml
C1 x 10ml = 0.15M x 50ml
C1 x 10 = 7.5
divide both side by 10
C1 = 0.75M
concentration of acid (HCl) is 0.75M
Answer:
B. They are stereoisomers
C. They are enantiomers
Explanation:
Let us consider all the options
A. D and L-glucose are not necessarily furanose, they can also be in free form (open chain) or as a six-membered ring (pyranose)
B. These sugars are stereoisomers as they have the same molecular formula, same bonds but with the different spacial arrangement.
C. Two structures are called enantiomers, if they are stereoisomers and are mirror images of each other and are not-superimposable. The given pair of structures satisfy these conditions
D. Epimers are diastereoisomers (same molecular formula and connectivity having a different spacial arrangement but are not mirror images and non-superimposable) with only one different stereocenter (if there are more than one). This is not the case
E. All monosaccharides (any sugar that cannot be hydrolysed to a simpler sugar) are reducing sugars. So, this option is invalid
