Answer:
Many things, but i think a toaster...
Explanation:
Answer:
At STP, 760mmHg or 1 atm and OK or 273 degrees celcius
Explanation:
The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.
This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable
Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas
Answer:
from the small intestine (RE = retinyl ester; ROH = ret- inol. CM = chylomicron remnant). is split at least one intact molecule of retinol, retinal.
Answer:
Rate of formation of SO₃ ![[\frac{d[SO_{3}] }{dt}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%5D)
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒ ![-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B2%7D%20%5D%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B3%7D%20%5D%7D%7Bdt%7D)
-----------------------------(1)
Given that the rate of disappearance of oxygen = ![-\frac{d[O_{2} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D)
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃ = ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒ ![\frac{d[SO_{3}] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D)
= 2 x 3.64 x 10⁻³ M/s
⇒ = 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃ = 7.28 x 10⁻³ M/s
If earth didn't rotate the global winds would stay on the same speed of rate because the wind isn't moving without rotation in the earths axis.
