Answer:
The four coefficients in order, separated by commas are 1, 8, 5, 6
Explanation:
We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.
C₅H₁₂ + ?O₂→ ?CO₂ + ?H₂O
We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.
As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.
Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.
We add 16 in the reactant side. Balanced equation is:
2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O
We also can divide by /2 in order to have the lowest stoichiometry
C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
, 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>
<u></u>
Answer: 26.5 mm Hg
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:

where,
= initial pressure at
= ?
= final pressure at
= 100 mm Hg
= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature =
Now put all the given values in this formula, we get
![\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BP_1%7D%7B100%7D%29%3D%5Cfrac%7B28000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B299.5%7D-%5Cfrac%7B1%7D%7B267.9%7D%5D)



Thus the vapor pressure of
in mmHg at 26.5 ∘C is 26.5