Answer: The density of Ammonia is 0.648 g/l
Explanation:
Density = Mass/ Volume
Mass of one mole of Ammonia (NH3) = 17.031g
Volume =?
Using the ideal gas law we can determine the volume.
PV = nRT
P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K
Make V the subject of the formular, we then have;
V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm
V = 24.04358/ 0.913 = 26.3L
Having gotten the value of Volume in this question, we then go back to solve for density.
Density = Mass/ Volume
17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l
Explanation:
The given data is as follows.
Now, according to Michaelis-Menten kinetics,
![V_{o} = V_{max} \times [\frac{S}{(S + Km)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%20V_%7Bmax%7D%20%5Ctimes%20%5B%5Cfrac%7BS%7D%7B%28S%20%2B%20Km%29%7D%5D)
where, S = substrate concentration = 
M
Now, putting the given values into the above formula as follows.
![V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%206.8%20%5Ctimes%2010%5E%7B-10%7D%20%5Cmu%20mol%2Fmin%20%5Ctimes%20%5B%5Cfrac%7B10.4%20%5Ctimes%2010%5E%7B-6%7D%20M%7D%7B%2810.4%20%5Ctimes%2010%5E%7B-6%7DM%20%2B%205.2%20%5Ctimes%2010%5E%7B-6%7D%20M%29%7D%5D)
= 
This means that 
would approache 
.
Since Na has a 1+ charge and O has a -2 charge, by reversing the charges and placing them as subscripts for the other atoms the formula is Na2O1 or simply Na2O.
I’m pretty sure it’s D.increases the activation energy for a reaction.
