Molar mass Argon = 39.948 g/mol
1 mol ------ 39.948 g
mol ----- 20.0 g
mol = 20.0 * 1 / 39.948
= 0.5006 moles
1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L
L = 0.5006 * 22.4
= 11.21 L
hope this helps!
Moles of Hydrogen present: 100 / 2 = 50 moles
Moles of Nitrogen present: 200 / 28 = 7.14 moles
Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles
Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.
Molar ratio of Nitrogen : Ammonia = 1 : 2
Moles of ammonia = 7.14 x 2 = 14.28 moles
Its a C with 4 dots, 1 on top,1 on bottom, 1 on the left and 1 on the right
For #4 is 298.48 hope it is correct
Answer:
The value of the Michaelis–Menten constant is 0.0111 mM.
Explanation:
Michaelis–Menten 's equation:
![v_o=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=v_o%3DV_%7Bmax%7D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D%3Dk_%7Bcat%7D%5BE_o%5D%5Ctimes%20%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
![V_{max}=k_{cat}[E_o]](https://tex.z-dn.net/?f=V_%7Bmax%7D%3Dk_%7Bcat%7D%5BE_o%5D)
Where:
= rate of formation of products
[S] = Concatenation of substrate
![[K_m]](https://tex.z-dn.net/?f=%5BK_m%5D)
= Michaelis constant
= Maximum rate achieved
= Catalytic rate of the system
![[E_o]](https://tex.z-dn.net/?f=%5BE_o%5D)
= Initial concentration of enzyme
On substituting all the given values
We have :
[S] = 0.10 mM
![\frac{v_o}{V_{max}}=\frac{[S]}{(K_m+[S])}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_o%7D%7BV_%7Bmax%7D%7D%3D%5Cfrac%7B%5BS%5D%7D%7B%28K_m%2B%5BS%5D%29%7D)
The value of the Michaelis–Menten constant is 0.0111 mM.
