The answer is 0.405 M/s
- (1/3) d[O2]/dt = 1/2 d[N2]/dt
- d[O2]/dt = 3/2 d[N2]/dt
- d[O2]/dt = 3/2 × 0.27
- d[O2]/dt = 0.405 mol L^(-1) s^(-1)
Answer:
both sugar and phosphate molecules
<span>Total mass = 2.75g
Mass % of Mg = 1.0g x 100/2.75g = 36.36 %
Mass % of O = 1.75g x 100/2.75g = 63.64 %
Mol of Mg = 36.36/24 = 1.515
Mol of O = 63.64/16 = 3.977
Ratio of Mol of Mg and O in the substance = (1.515 : 3.977) x 2 = 3 : 8
The empirical formula of substance is Mg3O8</span>
I know CL is chlorine and NA is sodium.
I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.
