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MXV= (0.194M)(1.00L)=0.194moles
42.39LiClg/molex0.194moles=8.2g LiCl
Given :
2NOBr(g) - -> 2NO(g) + Br2(g)
Initial pressure of NOBr , 1 atm .
At equilibrium, the partial pressure of NOBr is 0.82 atm.
To Find :
The equilibrium constant for the reaction .
Solution :
2NOBr(g) - -> 2NO(g) + Br2(g)
t=0 s 1 atm 0 0
1( 1-2x) 2x x
So ,
At equilibrium :
![K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cdfrac%7B%5BNO%5D%5E2%5Bbr_2%5D%7D%7B%5BNOBr%5D%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D%5Cdfrac%7B0.18%5E2%5Ctimes%200.9%7D%7B0.82%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D0.043%5C%20atm)
Hence , this is the required solution .
Answer:
Enthalpy of vaporization = 30.8 kj/mol
Explanation:
Given data:
Mass of benzene = 95.0 g
Heat evolved = 37.5 KJ
Enthalpy of vaporization = ?
Solution:
Molar mass of benzene = 78 g/mol
Number of moles = mass/ molar mass
Number of moles = 95 g/ 78 g/mol
Number of moles = 1.218 mol
Enthalpy of vaporization = 37.5 KJ/1.218 mol
Enthalpy of vaporization = 30.8 kj/mol
Answer:
The correct answer is : 0.025.
Explanation:
The precipitation reaction is as follows in this procedure:
NiSO4.6H20 (aq) + Na2CO3.10H2O (aq)⇒ NiCO3 (s) + Na2SO4 (aq) + 16H2O
So, the number of moles of reactants can be found by
number of moles : mass used/ molar mass
n (NiSO4×6H2O) = (6.57 g)/(262.84 g/mol) = 0.025 mol (for Ni)
n (Na2CO3×10H2O) = (7.15 g)/(286.14 g/mol) = 0.0250 mol (for CO3).
Thus, The maximum number of moles of NiCO3 that can form is 0.025 mol.
