Answer:
9.2
Explanation:
Let's do an equilibrium chart of this reaction:
2NO(g) + O₂(g) ⇄ 2NO₂(g)
4.9 atm 5.1 atm 0 Initial
-2x -x +2x Reacts (stoichiometry is 2:1:2)
4.9-2x 5.1-x 2x Equilibrium
The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:
y = 2x/(4.9 - 2x + 5.1 -x + 2x)
0.52 = 2x/(10 - x)
2x = 5.2 -0.52x
2.52x = 5.2
x = 2.06 atm
Thus, the partial pressure at equilibrium are:
pNO = 4.9 -2*2.06 = 0.78 atm
pO₂ = 5.1 - 2.06 = 3.04 atm
pNO₂ = 2*2.06 = 4.12 atm
Thus, the pressure equilibrium constant Kp is:
Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]
Kp = [(4.12)²]/[(0.78)²*3.04]
Kp = [16.9744]/[1.849536]
Kp = 9.2
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Markovnikov rule, in organic chemistry, a generalization, formulated by Vladimir Vasilyevich Markovnikov in 1869, stating that in addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component ...
Answer:- 1840 g.
Solution:- We have been given with 3.35 moles of and asked to calculate it's mass.
To convert the moles to grams we multiply the moles by the molar mass of the compound. Molar mass of the compound is the sum of atomic masses of all the atoms present in it.
molar mass of = atomic mass of Hg + 2(atomic mass of I) + 6(atomic mass of O)
= 200.59+2(126.90)+6(16.00)
= 200.59+253.80+96.00
= 550.39 gram per mol
Let's multiply the given moles by the molar mass:
= 1843.8 g
Since, there are three sig figs in the given moles of compound, we need to round the calculated my to three sig figs also. So, on rounding off to three sig figs the mass becomes 1840 g.
