Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
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Answer:
41264 g of CO₂
Explanation:
Combustion reaction is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 mol of propane react to 5 moles of oxygen in order to proudce 3 moles of carbon dioxide and 4 moles of water.
In a combustion reaction, our reactant reacts to oxygen and the products are always CO₂ and water.
We have the volume of propane but we need moles of it, so we need to apply density.
Density = mass / volume so mass = density . volume.
Density of propane is: 493 g/L
Mass of propane is 493 g/L . 27.9L = 13754.7 g
We convert mass to moles: 13754.7 g . 1 mol/ 44g = 312.6 moles
According to reaction, 1 mol of propane can produce 3 moles of CO₂
Our 312.6 moles will produce 312.6 . 3 = 937.8 moles
We convert moles to mass: 937.8 mol . 44 g/mol = 41264 g
The volume of nitrogen monoxide that occupy at STP is= 277 Ml
calculation
The volume is obtained using the combined gas law that is P1V1/T1= P2V2?
Where P1 = 720 MmHg
V1 = 400ml
T1= 100 +273= 373 K
At STP temperature = 273 K and pressure= 760 mm hg
therefore T2= 273 k
p2 = 760 mmhg
V2=?
make V2 the subject of the formula
V2= (T2 ×P1 ×V1)/(P2×T1)
V2 is therefore = (273k x720 mmhg x 400 ml)/(760 mmhg x373K) = 277 Ml
