Answer:
the mass of CaO present at equilibrium is, 0.01652g
Explanation:
![K_p = [CO_2]](https://tex.z-dn.net/?f=K_p%20%3D%20%5BCO_2%5D)
= 3.8×10⁻²
Now we have to calculate the moles of CO₂
Using ideal gas equation,
PV =nRT
P = pressure of gas = 3.8×10⁻²
T = temperature of gas = 1000 K
V = volume of gas = 0.638 L
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mole.k
Now we have to calculate the mass of CaO
mass = 2.95 * 10 ⁻⁴ × 56
= 0.01652g
Therefore,
the mass of CaO present at equilibrium is, 0.01652g
The balanced equation is
2SO
2
+
O
2
→
2SO
3
If your bike isn't the right size, your legs could be rubbing up against the seat of your bike. The friction can chafe the skin on your legs. If you were to crash your bike, skidding on the ground can cause anywhere from minor scrapes to serious friction burns.
<span>Molecular reaction
2HBr(aq) Ba(OH)2 (aq) = 2H2O (l) BaBr2(aq)
Ionic equation
2H 2Br- Ba 2 2OH- --------> Ba 2 2Br - 2H2O
Net ionic equation
2H (aq) 2OH- (aq)- ---------> 2H2O (l)</span>
This is where something called Graham's Law applies. Grahams law says that the rate of effusion is inversely proportional to the square root of the molar mass of a gas. What that means is that the less the molar mass of a gas is, the less quickly the gas effuses (effusion being the rate at which a gas can travel through a small hole). A great way to think about this is that effusion is thought about as though the gas is traveling through a small hole, so smaller particles would be able to go through it with greater ease than would a large particle. I don't know what particular sentences the question asks for are, but the answer should be that gas A (molar mass 4) has the greatest effusion rate, gas B (molar mass 16) has the second fastest effusion rate, and gas C (molar mass 32) has the slowest effusion rate.
