Answer:
A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
(4x^2)x
Kc= -----------
(A-2x)^2
PV=nRT
n/v = P/RT = .745/(0.0821)(334+273) = .01495
To Find the initial molarity of NO2
(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L
Thus:
46(A-2x) + 2x(30) + 32x = .515 g/L
46A-92x+60x+32x = .515
46A=.515
A=.01120 M
Using the total molarity found
(A-2x)+2x+x = .01495 M
A+x=.01495
Plug in A found into the above equation:
.01120+x = .01495
x=.00375
Now Plug A and x into the original Equilibrium Constant Expression:
(4x^2)x
Kc= -----------
(A-2x)^2
Kc = 0.000014
Explanation:
It’s about drive it’s about power
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
When the magma cools slowly. You find a deposit of organic limestone.
Answer: The moles of KIO cuboid titrated are 0.0002382
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in ml
Now put all the given values in the formula of molarity, we get
Therefore, the moles of KIO cuboid are 0.0002382
