Answer:
2,019 km
Explanation:
Step 1: Given data
Distance traveled by the car (D): 1,255 mi
Step 2: Convert the distance traveled by the car to kilometers
To convert one unit into another, we use a conversion factor. In this case, the appropriate conversion factor between miles and kilometers is 1 mile = 1.609 km. The distance traveled by the car, in kilometers, is:
D = 1,255 mi × (1.609 km/1 mi) = 2,019 km
Explanation:
Dichloromethane is flammable - FALSE
Methanol is flammable. - TRUE
Concentrated sulfuric acid is corrosive. - TRUE
10% sodium carbonate solution must be used in the fume hood. - FALSE
Benzoyl chloride is a lachrymator. - TRUE
Answer:
22.5mL is the volume of the water
Explanation:
When the graduated cylinder is in the 25.0mL mark, the mass of this volume is 22.4g. To convert this mass to volume we need to use density, as follows:
22.4g × (1mL / 0.99704g) = 22.5mL is the volume of the water.
That means the cylinder is uncalibrated in 2.5mL when the cylinder is in the 25.0mL mark
17,6% + 82,6% = 100%
mass of N = 14g
mass of H = 1g
17,6% H = 17,6g H
82,4% N = 82,4g N
17,6g : 1g = 17,6 moles of H
82,4g : 14g = 5,89 moles of N
N : H = 5,89 : 17,6 ||:5,89
N : H = 1 : 2,99≈3
the empirical formula of the compound = NH₃ (ammonia)
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.
