Answer:
m = 0.531 molal
Explanation:
∴ m fructose = 3.35 g
∴ V water = 35.0 mL
∴ ρ H2O = 1 g/mL
- molality = moles solute / Kg solvent
∴ Mw fructose = 180.16 g/mol
⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose
⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O
⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O
⇒ m = 0.531 molal
Answer:
Hot to cold
Explanation:
Unless people interfere, thermal energy or heat naturally flows in on direction only: hot towards cold. Heat moves naturally by any of three means. The processes are known as conduction, convection, radiation. Sometimes more than one may occur at the same time.
Answer:
Chemistry is the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.
They answer questions like, like "What makes up this material?," "How does this material change when heated or cooled,?" What happens when I mix this material with another material,?" and "What rules determine how materials change in different situations?" What makes up Chemical X?
Answer:
Option 2, Half of the active sites are occupied by substrate
Explanation:
Michaelis-Menten expression for enzyme catalysed equation is as follows:
![V_0=\frac{V_{max\ [S]}}{k_M+[S]}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%5C%20%5BS%5D%7D%7D%7Bk_M%2B%5BS%5D%7D)
Here, 
is Michaelis-Menten constant and [S] is substrate concentration.
When [S]=Km
Rearrange the above equation as follows:
![V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_M%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B%5BS%5D%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B2%5BS%5C%5C%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%7D%7B2%7D)
when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.
Therefore, the correct option is option 2.
