JOIN MY AMONG US GAME THE CODE IS QGORMF
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Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
= u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u
Answer : The equilibrium concentration of NO is, 0.0092 M.
Solution :
First we have to calculate the concentration of NO.
The given equilibrium reaction is,
Initially conc. 0 0 0.1576
At eqm. (x) (x) (0.1576-2x)
The expression of will be,
By solving the term, we get:
Neglecting the 0.0839 value of x because it can not be more than initial value.
Thus, the value of 'x' will be, 0.0742 M
Now we have to calculate the equilibrium concentration of NO.
Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M
Therefore, the equilibrium concentration of NO is, 0.0092 M.