The answer is B for the apex answer
I don’t actually know for sure, but I think it would sink. Most of the mass of the molecule is from the oxygen.
Amu of O2 = 32
Amu of C = 12
And since most of the balloon is oxygen, it has no reason to actually float. It would be denser than the air, I’d imagine. I could be totally wrong here, I’m guessing based on my knowledge of chemistry
The answer is B) fills all the space in its container
Answer:
a) K = [ CO2(g) ]
⇒ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
Explanation:
a) CaCO3(s) ↔ CaO(s) + CO2(g)
⇒ K = [ CO2(g) ]
∴ the [ CaCO3(s) ] does not appear in the denominator of the equilibrium constant, as it is a pure solid substance.
b) H2(g) + F2(g) ↔ 2 HF(g)
⇒ K = [ HF(g) ] ² / [ F2(g) ] * [ H2(g) ]
⇒ Kp = PHF² / PF2 * PH2
for ideal gas:
PV = RTn
⇒ P = n/V RT = [ ] RT
⇒ Kp = K (RT)∧Δn
⇒ the values of K and Kp are not the same.
c) K >> 1, The reaction has a high yield and is said to be shifted to the right. then the rate of the forward reaction is greater than the rate of the reverse reaction at equilibrium.
n = PV/RT
p = 1.6 atm
v = 12.7L
R = 0.0821
T = 24°C which is equivalent to 297.15 degrees k
n = (16 × 12.7) / (0.0821 × 297.15)
n = 20.32 / 24.39
n = 0.83 mol
C = 12.90
H = 1.0079
C2 = 12.010 × 2 = 24.02
H6 = 1.0079 × 6 = 6.0474
C2H6 = 30.0674
Ethane times n which is 30.0674 × 0.83mol
= 24.95 grams of C2H6. Which is Ethane.
